Download Ch. 27: Quantum Physics

course index
Recall from last lecture:
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Light is an electromagnetic wave created by oscillating electric and magnetic fields
The speed of light is c = 3.0×108 m/s.
These characteristics of electromagnetic waves (and light) result from the equations of electromagnetics
(Maxwell's equations).
CH. 26: RELATIVITY
The topics of the remaining chapters deal with modern physics, the term used to describe developments that have
occured since about 1896, when Roentgen discovered x-rays. The developments of modern physics include, special
and general relativity, an understanding of the structure of the atom, quantum mechanics, nuclear physics, particle
physics, astrophysics, transistors, and lasers.
Many popular books exist on these topics. Some recent ones accessible to the interested student are:
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A Brief History of Time by Stephen Hawking.
The Elegant Universe by Brian Green.
Hyperspace by Machio Iku.
The Jaguar and the Quark by .
The God Particle by Leon Lederman.
26.1 Introduction
Albert Einstein is irrevocably linked with relativity. In 1906 he published papers on the special theory of relativity,
the focus of most of this chapter, and ten years later on the general theory of relativity, which we will discuss briefly
at the end of this chapter. The special theory of relativity, or just relativity for short, knits space and time together into
the fabric of our existence. Basic beliefs about time and space must be reexamined in the context of relativity, as well
as our concepts of past and future. The general theory builds on the special theory, adding gravity to the mix,
resulting in a description of our world where gravity is just a manifestation of the geometry of space. Mass produces
a curvature of space, like the way a bowling ball will deform and curve a taught sheet.
Why does relativity force us to rethink our understanding of space and time, and does this mean that all the time spent
learning Newton's laws and their consequences was wasted?
I'll answer the second question first. No.
Well maybe a little more explanation is required. The effects of relativity manifest themselves when things move at
speeds near to the speed of light, 3×108m/s. Everyday objects on Earth move at speeds much less than the speed of
light, so that the difference between the results determined including the effects of relativity and without differ by
minuscule amounts. Said another way, the best way to solve problems of falling balls, masses on inclined planes, and
pendulums is exactly the way you learned.
Now to answer the first question. Relativity is based on two postulates:
1. The laws of physics are the same in all inertial reference systems.
2. The speed of light (in vacuum) is always measured to be 3×108m/s, independent of the motion of the observer
or of the source of light.
You used the first postulate throughout your study of mechanics. That is, Newton's laws apply equally well in a
smoothly flying airplane as on the surface of the earth. Note that an inertial reference system is one which is not
accelerating.
The second postulate is the new element. It's not too hard to see that if the speed of light is the same in all reference
systems, then the elapse of time will not be the same. Though it doesn't seem connected, this postulate will eventually
lead to that most famous of equations, E = mc².
26.2 The Principle of Relativity
In order to describe a physical event, it is necessary to choose a frame of reference. For example, for experiments
performed on the surface of the Earth can use the local surface as the reference frame. Someone passing by in a fast
moving vehicle might choose the vehicle as her reference frame. Would these two see a dramatic difference in an
event? Although they may not agree precisely on what occurred, they would both agree that whatever happened
followed Newton's laws.
To be more specific, consider a scenario with two observers, one on a fast moving jet aircraft, and the other on the
ground. Th observer on the airplane tosses a ball straight upward, and catches it when it falls back down. According
to him, the ball moves according to Newton's laws, rising and falling in gravity.
The observer on the ground sees the ball go up and down as well, but according to him, the ball is also moving
forward at the same speed as the aircraft, so it follows a parabolic path. Although the path reported is different, both
observers agree that the ball moves according to Newton's laws. This is the first postulate of relativity.
Example: C26.2
What two speed measurements will two observers in relative motion always agree on?
Both will always agree on the speed of light (in vacuum). It is always measured to be c. The second is their relative
speed. They must agree on their relative speed, otherwise their would be a difference between their inertial frames.
© Robert Harr 2000
course index
Recall from last lecture:
The two postulates of special relativity:
1. The laws of physics are the same in all inertial (non-accelerating) reference systems.
2. The speed of light (in vacuum) is always measured to be c = 3.0×108 m/s irrespective of the motion of the
observer or the source.
26.3 The Speed of Light
The second postulate of relativity is easy to state, but rather subtle in its implications. You see, never before had
something like the constancy of the speed of light occurred. In fact, after physicists realized that light is an
electromagnetic wave, they believed that there must be a medium in which the wave propagates. All other waves -sound, water, string vibrations -- propagated in a medium. So, went the thinking, there must be a medium, which they
called the ether, for the propagation of electromagnetic waves. If such an ether existed, then the speed of light would
equal c only when measured from a reference frame at rest relative to the ether. In reference frames that are moving
with respect to the ether, the speed of light could be greater than or less than c.
This situation is akin to an observer moving towards or away from a stationary source of sound. As seen in the
discussion of the Doppler effect in section 14.6, the speed of the observer adds or subtracts from the speed of sound
in air, resulting in a different relative speed. This results in a change in the number of cycles of the wave that reach
the observer in a given time, causing the frequency of the sound to change according to the Doppler formula.
26.4 The Michelson-Morley Experiment
The most famous experiment to detect the ether is the Michelson-Morley experiment. The experiment was performed
in Cleveland, at what is now Case Western Reserve University, and to get to the punch line, the result was negative,
no ether was detected.
The experiment is based on an interferometer, a device capable of detecting small changes along one light path
relative to a second light path. The change is detected by observing a shift of the interference pattern produced by
recombining the light from the two paths. In this interferometer, the two light paths are aligned perpendicularly. A
measurement is made by rotating the entire interferometer through 90° and watching for a small change due to the
change of orientation with respect to the velocity of the earth relative to the ether. No change of the interference
pattern is observed.
Michelson interferometers are once again at the forefront of Physics research. Two large interferometers are being
built for the LIGO experiment. LIGO stands for Laser Interferometer Gravitational-Wave Observatory. The facility
housing one of the interferometers is pictured. This facility is in Hanford, Washington, and the second is in
Livingston, Louisiana. In the photo you can see two long tubes emerging from the central building. Each tube is a 4
kilometer (2-½ mile) long vacuum for the light to travel in. Laser light is used for its superior properties in
interferometry.
As the name implies, the LIGO experiment will search for gravitational waves, a prediction of Einstein's general
theory of relativity which will be briefly discussed at the end of this chapter.
26.5 Einstein's Principle of Relativity
The null result of the Michelson-Morley experiment pretty much killed the ether hypothesis. (There were proposed
modifications to make the results consistent, but as Einstein pointed out, the simple conclusion is 'you don't need the
ether; it doesn't add anything to the Physics, so drop it'.
Maxwell's equations predict one speed for electromagnetic waves, and without an ether to provide a universal
reference frame, the speed of light has to be the same in all valid (that is, inertial) reference frames. In order to make
sense of the fact that everyone measures the same speed of light, we will have to alter our basic notions of space and
time.
Is it worth it?
I mean, isn't this stuff speculative, not well proven, and subject to change at almost any moment?
No. Special relativity is very well established and relied upon daily by millions of people around the world. The
Global Positioning System (GPS) satellites in orbit about the Earth carry very precise clocks; it is the precision of the
clocks that enable the GPS system to accurately determine your location. Effects from special relativity cause the
clocks to run slow by 7.11s per day. That's not much, but it must be corrected for the system to function properly.
One of the results of special relativity is that no object or particle can travel faster than the speed of light. In my
research work, we accelerate protons and electrons to extremely high speeds, nearly the speed of light, every day. The
protons are accelerated to a speed of 0.9999995c inside of a large ring of magnets. If special relativity were not
correct, the speed of the protons would reach 44 times the speed of light, and we wouldn't be able to follow them
properly around the ring -- distance traveled = speed × time would be 44 times greater!
Special relativity is now deeply ingrained in Physics and technology.
26.6 Consequences of Special Relativity
There are a number of consequences of special relativity, some of which you may be familiar with:
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Objects cannot travel faster than the speed of light.
Relativistic energy: E = mc².
Time and length are not absolute, invariant quantities.
Events that are simultaneous in one frame of reference may not be in another frame.
o Corollary: If event A occurs before event B in one frame of reference, it is possible that in another
frame of reference, event B occurs before event A.
We will begin by exploring the last two items in this list.
© Robert Harr 2000
course index
Recall from last lecture:
1. The laws of Physics are the same in all inertial reference frames.
2. The speed of light is always measured to be c = 3.0×108 m/s.
The Consequences of Special Relativity
Simultaneity and the Relativity of Time
Consider the rail car pictured in Figure 26.7, with two observers, one in the middle of the moving rail car at O', and
the other on the ground at location O. Lightning strikes both ends of the rail car, leaving marks at A, A', B, and B',
timed such that the observer on the ground at O sees the light from both strikes simultaneously.
Does the observer in the rail car see the strikes simultaneously? No. The rail car is moving forward at speed v. The
light from the lightning strikes propagates from A' and B' towards O'. Since O' is moving towards B', the light from
that end reaches the observer first; the light from the other end arrives a bit later.
The observer on the ground says that the lightning strikes were simultaneous, and the observer in the rail car says that
lightning hit at B' before A', and both are correct. In relativity, simultaneity is not absolute.
Note from this example that an observation is made when light (or some other signal) reaches an observer. This is an
important subtlety in relativity.
Time Dilation
To understand time dilation, let's do what Einstein called a "thought experiment". Imagine two observers, observer 1
inside a vehicle which is moving with respect to observer 2, as shown in Figure 26.8. Observer 1 makes a simple
clock by sending a pulse of light from a bulb, reflecting it from a mirror on the top of the vehicle, and measuring the
time when it returns to where it started. The time it takes for the light to travel from bulb to mirror to bulb is t1 = 2d
/ c. Think of this as one tick of this clock.
How long does it take for the light to travel from bulb to mirror to bulb for observer 2 who is outside the moving
vehicle? According to observer 2, the light leaves the bulb at an angle, travels up to the mirror, reflects at an angle,
and returns to the bulb. Let's say that the time it takes is t2. We can find t2 in terms of d and v, by considering the
path of the light as seen by observer 2. The path is a triangle, and if we divide it in half, we have two identical right
triangles (Figure 26.8c). The length of one hypotenuse is ct2/2, since it takes the light equal times to travel to the
mirror and back. The lengths of the sides are d, and vt2/2, the distance the vehicle moves in time t2/2. Using
Pythagorean's theorem:
(ct2/2)² = (vt2/2)² + d²
This can be solved for t2. First, move all the terms with t2 to the left side, and factor out t2²
(t2)²((c/2)² - (v/2)²) = d²
Next, divide both sides by ((c/2)² - (v/2)²), take the square root, and simplify to get:
t2 = 2d / Sqrt{c² - v²) = 2d / c Sqrt{1 - v²/c²}
Note that 2d / c = t1, so that we can relate the time of a tick of observer 1's clock to time measured by observer 2:
t2 = t1 / Sqrt{1 - v²/c²} = t1
where  = 1 / Sqrt{1 - v²/c²} is used to replace this commonly appearing term.
The  term is always greater than or equal to 1.  =1 only when the relative speed between two observers is 0. This
result means that observer 2 will say that observer 1's clock runs slow. If 1 second passes in observer 1's frame (t1 =
1sec.) then observer 2 will say that it actually took a time t2 = t_1 =  sec. The exact value of  depends on the
speed of observer 1 relative to observer 2:
description
satellite in geosynchronous orbit
half the speed of light
nine-tenths the speed of light
speed v

3
3.07×10 m/s = 11,000 km/hr 1.0000000001
0.5c
1.15
0.9c
2.29
ninety-nine-hundreths the speed of light 0.99c
7.09
Note that for a satellite in geosynchronous orbit, the difference one day is shorter than one day on earth by 4.3s.
The time that it takes for something to occur in a frame at rest with the action (the rest frame) is called the proper
time. Observer 1 measures the proper time of the light clock, but observer 2 is in motion relative to the clock and
therefore sees the effect of time dilation.
Example: P26.4
An astronaut at rest on Earth has a heartbeat rate of 70 beats/min. When the astronaut is traveling in a spaceship at
0.90c, what will this rate be as measured by (a) an observer also in the ship and (b) an observer at rest on the Earth?
(a) An observer in the ship is at rest relative to the astronaut, so their relative speed is 0. Thus the rate must be the
same. Note that for v=0, =1.
(b) The astronaut is moving with speed 0.90c relative to the observer on Earth. Thus, from the table above we find
that  = 2.29. Since one heartbeat takes (1/70)min = (6/7)sec = 0.857sec in proper time, then to the observer on Earth
it takes (2.29)(0.857sec) = 1.96 sec, and therefore her heartrate is 30 beats/min.
The following is a famous paradox of special relativity. This being a physics course, there is a definite answer and a
clear way to resolve the paradox, and that resolution may give you some additional insight.
The Twin Paradox
Consider this situation involving two 20 year old twin brothers, Speedo and Goslo. Goslo is a stay-at-home kinda guy
while Speedo signs up for a space trip to a planet 30 lightyears from Earth. Speedo takes off, accelerates to nearly the
speed of light, travels to the planet, then turns around, and comes back to Earth. On his return he finds that 60 years
have elapsed on Earth, yet he has aged just 10 years! While Goslo would claim that Speedo was traveling at a very
high relative speed, Speedo can equally well claim that he was at rest while Goslo and the rest of Earth was moving
at high speed relative to him. Given this symmetry, why is it that Speedo aged less?
The resolution is to note that there are times during Speedo's voyage when special relativity doesn't apply. For
instance, he must accelerate, then decelerate. He is the one who turns around after reaching the distant planet. Goslo,
and the rest of Earth continued on their normal routine during the entire time. This difference breaks the symmetry
mentioned above, and lets us differentiate Speedo's time in space from the rest of Earth.
Length Contraction
Not only does time change, as measured by observers in relative motion, but also distances. The proper length of an
object is its length measured in its rest frame. When the length is measured from a frame moving relative to the
object, the measurement is less than the proper length, hence the term length contraction.
An object of proper length Lp moving relative to an observer is measured by the observer to have a length:
L = Lp /  = L Sqrt{1 - v²/c²}
The contraction occurs only along the direction of motion. If the object is moving parallel to its length, then its width
and height remain unchanged. That is, if it is moving in the x direction, then the length is x is changed, while the
width in y and height in z remain the same.
© Robert Harr 2000
course index
Recall from last lecture:
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The rest frame is the reference frame at rest with an object or event.
Proper time and proper length are time and length measured in the rest frame.
Relativistic factor  = 1 / Sqrt{1 - v²/c²}
Time dilation: t2 = t1
Length contraction: L = Lp /  = Lp Sqrt{1 - v²/c²}
Example: P26.8
A friend in a spaceship travels past you at a high speed. He tells you that his ship is 20m long and that the identical
ship that you are sitting in is 19m long. According to your observations, (a) how long is your ship, (b) how long is his
ship, and (c) what is the speed of your friend's ship (relative to yours)?
(a) Since your ship is identical to your friend's, it must have the same proper length. The proper length of his ship is
20m, so that must be the proper length of your ship, and this is what you would measure while in the rest frame of
your ship.
Your friend, moving with speed v relative to you, measures your 20m ship to be 19m long. You are moving with
speed v relative to your friend, so you will measure his 20m long ship to be 19m long as well.
To determine the relative speed of the ships, use the length contraction formula
L = Lp Sqrt{1 - v²/c²},
so (1 - v²/c²) = (L/Lp)².
Move things around a bit more to get that
v²/c² = 1 - L²/Lp²
or v/c = Sqrt{1 - L²/Lp² = Sqrt{1 - (19/20)²} = 0.22. Therefore, we find that v = 0.22c.
Note in the last example that it is perfectly reasonable to leave the answer for a speed as a fraction times the speed of
light.
26.7 Relativistic Momentum
Recall that in "classical" mechanics, momentum was useful because it is conserved in many situations. We want to
modify the definition of momentum for special relativity so that it is also conserved in situations, such as in
collisions. We also want this new definition to reduce to the old definition of momentum when the speed approaches
zero. The relativistic quantity that satisfies these conditions is
p = mv / Sqrt{1 - v²/c²} = mv
As we saw earlier, when the speed is small, the factor  is nearly equal to 1, and then this expression reduces to the
familiar non-relativistic expression p=mv.
Example: P26.18
An unstable particle at rest breaks up into two fragments of unequal mass. The mass of the lighter fragment is
2.50×10-28kg (a pion), and that of the heavier fragment is 1.67×10-27kg (a proton). If the lighter fragment has a speed
of 0.893c after the breakup, what is the speed of the heavier fragment?
Relativistic momentum is conserved, just as in the non-relativistic case. The initial momentum of the system is zero
since the unstable particle is at rest. For the final momentum to be zero, then the momenta of the two particles must
have equal magnitude and be in opposite directions (sketch).
l = 1 / Sqrt{1 - 0.893²} = 2.22
The momentum of the light particle is
pl = l ml vl
and this must equal the momentum of the heavy particle,
ph = h mh vh.
Therefore,
h vh = l vl (ml / mh) = (2.22)(0.893c)(2.50×10-28kg / 1.67×10-27kg) = 0.297c
Now, it takes a bit of algebra to solve this for v, since v appears in a square root in the  term. The solution works out
to be
vh = c / Sqrt{(1/0.297)² + 1} = 0.285c.
26.8 Relativistic Addition of Velocities
Earlier we discussed that if two objects (spaceship 1 and spaceship 2) are moving with respect to each other, then the
velocity of 2 measured with respect to 1, v21, must equal the velocity of 1 measured with respect to 2, v12.
v21 = v12
Suppose there is a third object (rocket 3), and it is moving with respect to 2 with velocity v23, in the same direction as
v12. What is the velocity of 3 with respect to 1, v13?
As you might now expect, the result is not a simple addition of velocities. Instead, we "add" velocities according to
the following equation:
v13 = (v12 + v23) / (1 + (v12 v23 / c²))
To see what the consequence of this equation are, let's look at an example.
Example: P26.22
A space vehicle is moving at a speed of 0.75c with respect to Earth. An atomic particle is projected at 0.90c in the
same direction as the spaceship's velocity with respect to an observer inside the vehicle. What is the speed of the
projectile as seen by an observer on Earth?
Let's identify the Earth as object 1, the vehicle as object 2, and the projectile as object 3. Therefore, v12 = 0.75c, and
v23 = 0.90c. Note that v23 is positive since the particle moves in the same direction as the vehicle. Adding the
velocities as learned previously, we would get the result that the speed of the projectile as seen from Earth is v13 = v12
+ v23 = 0.75c + 0.90c = 1.65c! This is greater than the speed of light which is not possible! Using the relativistic
equation we get:
v13 = (0.75c + 0.90c) / (1 + (0.75c)(0.90c)/c²) = 1.65c / 1.675 = 0.985c
This value is close to the speed of light, but not greater!
The relativisitic addition of velocities formula always yields a result less than the speed of light.
© Robert Harr 2000
course index
SI session: Are there people interested in attending, but find that the present times are inconvenient. This week the
Friday sessions are cancelled.
Recall from last lecture:
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 = 1 / Sqrt{1 - v²/c²}
Time Dilation: t2 = t2
Length Contraction: L = Lp /  = Lp Sqrt{1 - v²/c²}
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Relativistic Momentum: p = mv
Relativistic Addition of Velocities: v13 = (v12 + v23) / (1 + (v12 v23 / c²))
26.9 Relativistic Energy
In relativity, we must modify the way we calculate kinetic energy for reasons similar to those that justify changing
the calculation of momentum. The relativistic kinetic energy of an object is given by:
KE = mc² - mc²
This expression gives nearly the same result as KE = ½mv² for speeds that are small compared to the speed of light,
but it also gives the correct result for large speeds.
The constant term mc² is called the rest energy of the object,
ER = mc².
The total energy of an object is its kinetic energy plus rest energy:
E = KE + mc² = mc² = mc² / Sqrt{1 - v²/c²}
This is the general form of the relativistic energy equation. You obtain the familiar E=mc² expression when the speed
of the object is zero, and =1.
Example: Rest Energy of Electron and Proton
Using E = mc² we can relate mass to energy. When discussing atoms, nuclei, and particles, it is common to use the
rest energy for particles in eV (or keV or MeV) rather than their mass in kg. Determine the rest energies (rest mass)
of the electron and proton in MeV?
The mass of the electron and proton are me = 9.11×10-31kg and mp = 1.67×10-27kg. Using E=mc² will yield the energy
in joules, which can be converted to electron-volts (eV) by the conversion factor (1 eV / 1.60×10-19J). For the
electron:
Ee = (9.11×10-31kg)(3.00×108m/s)2 (1 eV / 1.60×10-19J) = 5.11×105eV = 0.511MeV
and for the proton:
Ep = (1.67×10-27kg)(3.00×108m/s)2 (1 eV / 1.60×10-19J) = 9.38×108eV = 938MeV.
Example: Proton's Speed
At the accelerator where I perform research, protons are accelerated to an energy of 920GeV. What is the speed of
these protons?
These protons have a relativistic energy of
E = mp c² = Ep
where Ep is the rest energy of the proton. From this we find that these  for these protons is:
 = E / Ep = 920GeV/938MeV = 991.
Now, we can solve for v since  = 1/Sqrt{1 - v²/c²}:
2 = 1 / (1 - v²/c²)
1 - v²/c² = 1/2
v²/c² = 1 - 1/2 = (2 - 1)/2
v/c = Sqrt(2 - 1) /  = Sqrt(9912 - 1) / 991 = 0.9999995
Example: P26.28
The sun radiates approximately 4.0×1026J of energy into space each second. (a) How much mass is converted into
energy of other forms each second? (b) If the mass of the Sun is 2.0×1030kg, how long can the Sun survive if the
energy transformation continues at the present rate?
(a) The energy radiated by the sun is derived primarily from converting mass into other forms of energy. The energy
radiated is related to converted mass by E = mc². Therfore:
m = E / c² = (4.0×1026J) / (3.0×108m/s)2 = 4.4×109kg
(b) The above amount of mass is "burned" every second, so the total time that this can continue is:
t = (2.0×1030kg) / (4.4×109kg/s) = 4.5×1020s = 1.4×1013yrs = 14 trillion years.
26.10 General Relativity
This section is optional; this discussion is made to connect topics of popular science presentations to material in this
course.
Einstein felt that his "most pleasant thought" was the realization of the "equivalence principle". The equivalence
principle is the basis of his theory of general relativity. It states that a gravitational field is equivalent to a local
acceleration.
To understand what this means, consider the following thought experiment (sketch). A Physics 2140 student is in an
elevator, and drops a ball, noting that the ball falls to the floor of the elevator according to y = y0 - ½gt². This will be
exactly the result if the elevator is on Earth, and stationary.
What would happen if instead the elevator were on a space ship, in a gravity free region of space, and accelerating in
the upward direction with acceleration a = g? In this case, if a ball is dropped, it continues moving with the same
instantaneous velocity of the elevator at the moment it is released. Because it is no longer attached to the elevator, it
stops accelerating. But the elevator continues to accelerate, so while the position of the ball can be described by yball =
y0 + v0t, the position of the floor of the elevator is given by yfloor = v0t + ½at², with a = g. So the distance of the ball
from the floor is given by:
yball - yfloor = y0 - ½gt²
This is identical to the result if the elevator is on Earth in a gravitational field, g.
In fact, no experiment that we do can differentiate between a local gravitational field, and a local acceleration. Thus
the equivalence between acceleration and gravity.
Einstein took this idea and developed it into the general theory of relativity, with some remarkable predictions. The
first is that, with gravity replaced by acceleration, what we call the gravitational force comes from local curvatures of
space. We can visualize this in two dimensions with a stretched out sheet to represent two dimensions of space. The
presence of mass (like a star) can be simulated by placing a weight on the sheet. The weight causes the sheet to curve
in its vicinity, similar to the way that a star curves space in its vicinity. If we roll a marble across the sheet in the
region where it is curved, the marble will deflect. In general relativity, the curvature of space causes the path of
planets to curve, resulting in what we call gravity!
The next surprising result is that the same rules apply to light. That is, light will also curve when passing near a
massive object. The curvature is not great, but in recent years it has been clearly seen by many telescopes. (Pictures
of gravitational lensing.)
A third result is that clocks run more slowly where gravity is larger. This again has been seen, and recalling the
earlier discussion of the global positioning satelltes, the correction of their clocks due to general relativity is even
larger than the correction for special relativity. The special relativity effects cause the clocks to run slow by 7.11s,
and the general relativity effects cause them to run fast by 45.7s per day!
This is also the cause of gravitational red shifting. The light from massive stars is shifted in frequency from blue to
red as the light emerges from the gravitational field of the star.
A fourth prediction is that if the mass of a star becomes sufficiently great, it will curve space to such an extent that
even light cannot escape from the star. The resulting object is called a black hole. The evidence for the existence of
black holes is now quite strong. It is assumed that nearly every galaxy has a black hole at its center, and that objects
known as quasars are very distant black holes. But as you might guess, since black holes don't radiate like stars, the
evidence for their existence is indirect.
Finally, general relativity predicts the existence of gravity waves. Gravity waves cause local distortions in space as
they propagate, and measurable waves are presumed to be produced in only the most cataclysmic events, such a black
hole swallowing a star, or the collision of two black holes. Physicists have been trying to detect gravity waves for
over 30 years, but they are very feeble, requiring extremely sensitive equipment. The LIGO experiment is the latest
attempt; its state of the art interferometer may reach sufficient sensitivity to see the first signals of gravity waves in
the next five years.
© Robert Harr 2000
course index
CH. 27: QUANTUM PHYSICS
The theories of special and general relativity represent on of two major developments in Physics in the early 20th
century. The second major development is quantum theory.
While special relativity is important for the study objects that move at close to the speed of light, and general
relativity is important for the study of very massive objects, quantum theory is important for the study of very small
objects (molecules, atoms, nuclei, and particles) or processes where small amounts of energy are transferred
(transistors, superconductors, and blackbody radiation).
Recall that the expressions of special relativity reduce to the traditional expressions in the case of small velocities.
This was required so that the results of "classical" or "Newtonian" Physics are reproduced. The same reduction holds
for quantum Physics -- results from quantum theory must reduce to the "classical" Physics results as the objects
involved become large (more massive) or the energy exchanged becomes large.
27.1 Blackbody Radiation and Planck's Hypothesis
We traditionally begin discussing quantum Physics with the rather mundane sounding topic of blackbody radiation.
Historically this topic marks the introduction of quantized energy, the basis of quantum theory.
A blackbody is something that absorbs all the light incident on it. The concept of a blackbody is introduced in section
11.7 of the text, in a discussion of heat transfer by radiation. The basic idea is that a blackbody at temperature T emits
heat in the form of radiation according to Stefan's law:
P = A e T4
where A is the surface are of the body, e is emissivity of the surface (e=1 for a blackbody), T is the temperature in
degrees kelvin, and  is Stefan's constant,  = 5.6696×10-8W/m2K4.
With the understanding that light is electromagnetic radiation, physicists began investigating the properties of
blackbody radiation. For experimenters, a blackbody is constructed by a hollow cavity with a small hole from which
the radiation can emerge. I tend to think of this like a hot oven as used in glass blowing. It was discovered that the
radiation is emitted with a particular spectrum which only depends on the temperature of the blackbody. That is, the
different wavelengths of infrared, visible, and ultraviolet radiation come in proportions that only depend on the
temperature. (drawing of spectrum) The peak of the spectrum is given by Wien's displacement law:
max T = 0.2898×10-2 m.K
Attempts to explain the spectrum using classical physics failed. They always tended to predict too much radiation
with small wavelengths, hence known by the term "ultraviolet catastrophe".
Max Planck came up with a solution to the problem that involved two assumptions:
1. The radiation inside the blackbody is absorbed and re-emitted from the walls. The wall is composed of
"resonators" which can have only certain discrete energy values, En = nhf, where n is a positive integer called
the "quantum number", f is the frequency of the resonator, and h is Planck's constant:
h = 6.626×10-34J s
We say that the energy of the resonators is "quantized".
2. The second assumption is that the electromagnetic radiation is also quantized, it comes in packets called
photons. When radiation is absorbed or emitted, the resonator changes from one quantum state to another; if
the resonater does not change state, no energy is emitted or absorbed. If the resonater changes from the state
with n=3 to the state with n=2, then a photon is emitted with energy
E = 3hf - 2hf = hf.
With the addition of these assumptions, Planck was able to explain the observed blackbody spectrum exactly.
In recent years, the importance of this mundane topic has grown and flourished. The reason is that space, the
universe, acts like a blackbody (at least in regions away from the sun, planets, stars and galaxies). The blackbody
radiation of the universe was discovered in the 1940's when some Bell Labs researchers were trying to set up a
microwave receiver. Their receiver was picking up noise that was soon realized to be radiation from the "cosmic
microwave background" (CMB).
The cosmic microwave background is a remnant of the "big bang". It has more recently been measured with
incredible accuracy by experiments such as the COBE satellite. Their result is pictured below. The shape of this
spectrum is exactly as predicted for a black body of temperature T = 2.782°K.
Example: P27.2
A certain light source is found to emit radiation whose peak value has a frequency of 1.00×1015Hz. Find the
temperature of the source assuming that it is a blackbody radiator.
We want to use T = (0.2898×10-2 m.K) / max, but what we are given is the frequency of the maximum, fmax =
1.00×1015Hz. Recall that for electromagnetic radiation, frequency and wavelength are related by
c=f
Therefore,
max = c / fmax = (3.00×108m/s) / (1.00×1015Hz) = 3.00×10-7m
and
T = (0.2898×10-2 m.K) / (3.00×10-7m) = 9660°K
Example: P27.6
A quantum of electromagnetic radiation has an energy of 2.0keV. What is its wavelength?
Again use the fact that c = f, and the energy of a photon is related to its frequency by E = hf to find that
 = hc/E = (6.63×10-34Js)(3.0×108m/s) / (2.0×103eV)(1.6×10-19J/eV) = 6.2×10-10m = 0.62nm
This corresponds to an x-ray.
© Robert Harr 2000
course index
Recall from last lecture:





c = f for electromagnetic radiation
Peak of blackbody spectrum: maxT = 0.2898×10-2m.K
Electromagnetic radiation quantized in packets called photons.
Planck's constant h = 6.63×10-34Js
The energy of a photon of frequency f is Eph = hf.
27.2 The Photoelectric Effect
Although Planck's resolution of the "ultraviolet catastrophe" for the blackbody radiation spectrum assumed that
electromagnetic radiation is quantized in packets of energy we call photons, it was not understood if this was reality,
or simply a coincidence of nature. The photoelectric effect was the first clear demonstration that photons are real, not
just a fudge.
The photoelectric effect occurs when light (visible or ultraviolet) shines on a metal. Energy from the light can cause
electrons to be ejected from the surface of the metal. Let's review the reasons why the photoelectric effect supports
the postulate of quantization of electromagnetic energy in photons.
The photoelectric effect experiments use a device called a photocell. A photocell is made of two "electrodes" (pieces
of metal) inside of an evacuated glass bulb. (drawing like Figure 27.4) One of the electrodes is prepared for the
photoelectric effect, and is called the photocathode (a cathode is a source of electrons). The second electrode doesn't
participate in the photoelectric effect, but is intended to collect the ejected electrons, and is called the anode.
When light shines on the photocathode, electrons are ejected, and a current is produced. A potential can be applied
between the electrodes to enhance or limit the current. The photocell acts like a switch that is turned on or off by
light.
If light is a continuous wave instead of being quantized in photons, then how would we expect a photocell to behave.



We would expect that the rate of electrons emitted depends on the frequency and intensity of the light.
If no current was present for a particular frequency and intensity, then increasing the intensity or the voltage
between the electrodes should result in a current.
The kinetic energy of the ejected electrons should increase if the intensity of the light is increased.
The experiments observe different behavior.





The rate of electrons emitted does depend on the intensity and frequency of the light.
No current is present if the frequency is less than a minimum value called the cutoff frequency, fc.
For a given frequency, there is a negative voltage (the stopping potential, Vs, between the electrodes that
stops all current flow.
For a given frequency and voltage setting, the electrons have a maximum kinetic energy that doesn't depend
on the intensity of the light. The kinetic energy is related to the stopping potential, KEmax = e Vs.
The time delay between the incident light and the emitted electrons is very short, about a nanosecond, and
doesn't depend on the intensity.
These observations convinced physicists of the existence of photons.
Einstein is the one who explained the observations by suggesting that electromagnetic radiation is quantized in
photons. According to Einstein, the energy of the photon, hf, is transferred to a single electron. There's a certain
minimum amount of energy required for the electron to leave the surface of the metal, known as the work function, .
The resulting maximum kinetic energy of the electron is the difference between the photon energy and the work
function:
KEmax = hf - 
If the frequency is less than the cutoff frequency, then hf < , and there is not sufficient energy for the electron to be
ejected from the metal. The cut off frequency is given by hfc = . We can use the relation between frequency and
wavelength to find the cutoff wavelength:
c = c/fc = hc/
Example P27.14
Consider the metals lithium, aluminum, and mercury, which have work functions of 2.3eV, 4.1eV, and 4.5eV,
respectively. If light of wavelength 3.0×10-7m is incident on each of these metals, determine (a) which metals exhibit
the photoelectric effect and (b) the maximum kinetic energy for the photoelectrons for those that exhibit the effect.
(a) Begin by calculating the energy in eV of photons of light of the given wavelength.
Eph = hf = hc/ = (6.63×10-34Js)(3.0×108m/s) / (3.0×10-7m)(1.6×10-19J/eV) = 4.1eV.
The photoelectric effect is exhibited when the photon energy is greater than the work function. This is true for lithium
(2.3eV), but not so for aluminum (4.1eV) and mercury (4.5eV).
(b) For lithium, KEmax = hf -  = 4.1eV - 2.3eV = 1.8eV.
27.3 Applications of the Photoelectric Effect
There are many applications of the photoelectric effect. Most of them apply a photocell to a problem -- controlling
street lights, the breathalyzer, and electric eyes. Many of these have replaced the photocell with the more modern,
solid state photodiode. The device may have changed, but the idea is the same.
Many modern experiments use a variant of the photocell which is capable of detecting single photons, the
photomultiplier tube. For instance, biological studies which rely on measuring the flouresence of a tracer molecule
generally use photomultiplier tubes to measure the small number of flouresence photons.
27.4 X-rays
In 1895, Wilhelm Roentgen discovered x-rays, thus marking the beginning of modern physics. X-rays are a form of
electromagnetic radiation, like radio waves and light, but having higher frequency and energy than either. The energy
of x-rays is higher than the electron binding energies of light atoms. This enables x-rays to easily pass through water
and tissues, composed mostly of hydrogen, carbon, nitrogen, and oxygen, while they will tend to scatter off of bones
with their high density of calcium.
When originally discovered, it wasn't clear what x-rays where, hence the reason behind the name. It was easily
verified that x-rays are not deflected by electric or magnetic fields, and therefore are not charged particles.
It took a demonstration of diffraction of x-rays by a crystal, from which their wavelength could be calculated, and
conclude that they are electromagnetic waves.
A common way to produce x-rays is to rapidly decelerate high energy electrons, for instance by having them hit a
target made of heavy atoms. An x-ray tube is a device that does just this. The resulting x-rays have a broad spectrum
of energies, plus some sharp lines corresponding to atomic energy levels of inner shell electrons.
The x-ray is created when the electron passes near the nucleus of an atom and feels its electric field. This process is
like a collision of billiard balls, or the inverse of the photoelectric effect. The maximum x-ray energy, and minimum
wavelength results when the electron loses all its energy in a single collision, such that
eV = hfmax = hc/min
or therefore
min = hc/eV
Example: P27.21
What minimum accelerating voltage would be required to produce an x-ray with a wavelength of 0.0300nm?
Since min = hc/eV, we can turn this around to find
Vmin = hc/e = (6.63×10-34Js)(3.0×108m/s)/(1.6×10-19C)(3.00×10-11m) = 4.1×104V = 41kV.
27.5 Diffraction of X-rays by Crystals
27.6 The Compton Effect
© Robert Harr 2000
course index
Recall from last lecture:
27.4 X-rays
In 1895, Wilhelm Roentgen discovered x-rays, thus marking the beginning of modern physics. X-rays are a form of
electromagnetic radiation, like radio waves and light, but having higher frequency and energy than either. The energy
of x-rays is higher than the electron binding energies of light atoms. This enables x-rays to easily pass through water
and tissues, composed mostly of hydrogen, carbon, nitrogen, and oxygen, while they will tend to scatter off of bones
with their high density of calcium.
When originally discovered, it wasn't clear what x-rays where, hence the reason behind the name. It was easily
verified that x-rays are not deflected by electric or magnetic fields, and therefore are not charged particles.
It took a demonstration of diffraction of x-rays by a crystal, from which their wavelength could be calculated, and
conclude that they are electromagnetic waves.
A common way to produce x-rays is to rapidly decelerate high energy electrons, for instance by having them hit a
target made of heavy atoms. An x-ray tube is a device that does just this. The resulting x-rays have a broad spectrum
of energies, plus some sharp lines corresponding to atomic energy levels of inner shell electrons.
The x-ray is created when the electron passes near the nucleus of an atom and feels its electric field. This process is
like a collision of billiard balls, or the inverse of the photoelectric effect. The maximum x-ray energy, and minimum
wavelength results when the electron loses all its energy in a single collision, such that
eV = hfmax = hc/min
or therefore
min = hc/eV
Example: P27.21
What minimum accelerating voltage would be required to produce an x-ray with a wavelength of 0.0300nm?
Since min = hc/eV, we can turn this around to find
Vmin = hc/e = (6.63×10-34Js)(3.0×108m/s)/(1.6×10-19C)(3.00×10-11m) = 4.1×104V = 41kV.
27.5 Diffraction of X-rays by Crystals
Diffraction occurs when light passes through a regularly spaced set of slits. This week, you are using a diffraction
grating made by cutting 300 grooves per millimeter in a piece of glass -- grooves work just as well as slits for this
work. For a diffraction grating to work well, the spacing of the slits must be comparable to the wavelength of the
radiation. Visible light has wavelengths between about 700nm (red) and 400nm (blue). A diffraction grating with 300
grooves per mm has a slit spacing of (1/300)mm = 3m = 3000nm. This is 4 to 8 times larger than the wavelengths of
visible light, sufficient for your laboratory exercise.
The wavelengths of x-rays are about 0.1nm. Your diffraction grating will have little effect on x-rays. But it was
realized by Max von Laue that the wavelength of x-rays is comparable to the size of an atom (0.1nm = 1 angstrom),
and that a material with a regularly spaced grid of atoms could diffract x-rays, assuming that x-rays are
electromagnetic radiation. X-rays are EM radiation, they are diffracted by crystals, and Laue won a Nobel Prize.
Diffraction of x-rays by crystals continues to be an important research tool today. However, the crystals are usually
made from proteins, and the technique is called protein crystallography. The diffraction pattern produced is vital in
deducing the structure of proteins (folding), enabling researchers to connect the functionality of the protein with the
structure. Pharmaceutical researchers are common users of this technique.
27.6 The Compton Effect
The Compton effect is another famous experiment supporting the existence of photons. The Compton effect is the
scattering of x-rays from electrons (and for this reason is also called Compton scattering). To understand Compton
scattering, we must not only treat EM radiation as quantized photons, we must think of the photons as particles! After
all, how can a wave scatter off a point-like particle? Does sound scatter off a pencil? The only way to make sense of
Compton scattering, is to treat the photon like a particle not a wave.
Let's do just that and see how this works. Imagine an incoming photon scattering off a stationary electron, like
billiard balls colliding on a pool table (drawing, Figure 27.17). After they collide, the photon and electron will move
away at some angle to the original photon direction. If we know the energy (momentum) of the incoming photon, and
the outgoing angle of the photon, then we can determine the outgoing photon's energy (and the outgoing angle and
momentum of the electron) by requiring that energy and momentum be conserved. We treat the photon like a particle
with zero mass, and momentum p = E/c. (Note that physicists are more willing to call a wave a particle than to give
up energy and momentum conservation.)
It is traditional to express this in terms of the change of the photon's wavelength, since energy and wavelength are
related by E = hf = hc/. The result is:
 =  - 0 = (h/mec)(1 - cos)
where 0 is the wavelength of the incoming x-ray,  is the wavelength of the outgoing x-ray,  is the scattered angle
of the outgoing x-ray, and (h/mec) = 0.00243nm is called the Compton wavelength. The Compton wavelength is the
amount by which the photon's wavelength changes when it scatters at 90°. The change in wavelegth is smaller if the
scattering angle is less than 90°, and is a maximum of twice the Compton wavelength if the photon scatters by 180°
(scatters backwards). Since the Compton wavelength is small, the change in wavelength is only perceptible for
photons of small wavelength, say less than about 0.1nm, corresponding to x-rays.
Example: P27.29
X-rays with an energy of 300keV undergo Compton scattering from a target. If the scattered rays are deflected at 37°
relative to the direction of the incident rays, find (a) the Compton shift at this angle, (b) the energy of the scattered xray, and (c) the kinetic energy of the recoiling electron.
(a) The "Compton shift" is the change of wavelength, and for a 37° scattering angle this is:
 = (h/mec)(1 - cos) = (0.00243nm)(1 - cos37°) = 0.00049nm.
(b) We will need the quantity hc several times, so let's calculate it, and convert J to eV.
hc = (6.63×10-34Js)(3.0×108m/s) = 2.0×10-25Jm
hc = (2.0×10-25Jm)/(1.6×10-19J/eV) = 1.24×10-6eV m
The wavelength of the incoming x-ray is
0 = hc/E0 = (1.24×10-6eV m) / (3.00×105eV) = 4.13×10-12m = 0.00413nm
The outgoing wavelength is
 = 0 +  = 0.00413nm + 0.00049nm = 0.00462nm.
Photons of this wavelength have energy:
E = hc/ = (1.24×10-6eV m) / (4.62×10-12m) = 0.268×106eV = 268keV.
We can determine the kinetic energy of the recoiling electron because kinetic energy is conserved. The energy before
the scattering is the energy of the incoming x-ray, plus the rest energy of the electron. The energy after the scattering
is the energy of the outgoing x-ray, plus the kinetic energy of the recoiling electron, plus the rest energy of the
electron. These must be equal, therefore E0 + mec2 = E + KE + mec2, or
KE = E0 - E = 300keV - 268 keV = 32keV.
© Robert Harr 2000
course index
Recall from last lectures:









Blackbody radiation: Ad hoc quantization of energy of "radiators" and electromagnetic radiation gives
predictions that agree with observed radiation spectrum, maxT = 0.2898×10-2m.°K.
Introduced Planck's constant h = 6.63×10-34Js.
Photoelectric effect: Electromagnetic radiation quantized in packets, photons, of energy E = hf.
KEmax = hf - .
X-rays: radiation created by stopping energetic electrons with heavy atoms, min = hc / eV.
X-ray diffraction: X-ray photons behave like waves!
The Compton effect: X-ray photons behave like particles! (at least under certain circumstances).
 =  - 0 = (h/mec) (1 - cos).
27.7 Pair Production and Annihilation
Thus far, we've discussed processes that show that
1. quantized energy is needed to explain the blackbody spectrum,
2. electromagnetic radiation is quantized in photons, and
3. photons behave like particles under certain conditions (scattering).
The processes of pair production and annihilation are dramatic demonstrations of the equivalence of mass and
energy.
Pair production occurs when the energy of a very high energy photon is converted into mass, specifically an electron
and its anti-particle, a positron (sketch a diagram, like Figure 27.18). The reason why an electron and a positron are
produced is to conserve charge -- the photon is neutral, so the pair of particles must also be neutral. There are other
conservation laws that force the positive particle to be a positron, rather than, say, a proton. (An anti-particle is
identical to its particle except for the sign of the charge.) This process must also conserve energy and momentum. It
turns out that a nearby nucleus is required to allow momentum to be conserved. Conservation of energy then requires
that the energy of the photon be greater than the rest energy of the electron and positron:
hfmin = 2 me c2 = 2 (0.511MeV) = 1.02MeV
Photons of this high energy are generally referred to as gamma rays, rather than x-rays.
Pair annihilation is the process where an electron and positron combine, converting their mass into energy (photons).
We can generally assume that the electron and positron are at rest. Again, momentum and energy are conserved. The
initial momentum is zero if the electron and positron are at rest, so after annihilation, two photons are created,
moving in opposite directions. The photons have equal momentum and energy. The sum of the photon energies
equals the sum of the electron and positron rest mass energy, or
Eph = me c2 = 0.511MeV.
The unique photon energy is a signature of pair annihilation. Pair annihilation is appropriated for medical use in
positron emission tomography (PET), a procedure sometimes used to locate tumors, especially in the brain.
Example: P27.32
How much total kinetic energy will an electron-positron pair have if produced by a photon of energy 3.00MeV?
Any energy beyond the minimum needed to produce the electron and positron goes into kinetic energy. That is
Eph = 2me c2 + K.E.
thus, in this case,
K.E. = 3.00MeV - 1.02MeV = 1.98MeV.
27.8 Photons and Electromagnetic Waves
How can photons act like waves and particles?! You will probably believe me when I tell you that they can show
properties of each, and wonder why physicists make such a big deal about it. For our purposes, I'd like to explore this
point a little deeper.
Particles are rather familiar by now. Particles have a definite location in space, and may have a physical size, or be
infinitismal, like geometric points. Usually the particles we deal with have mass, but let's relax this requirement for
photons, since they have no mass. When they move, particles follow a definite path, a curve through space.
Waves have been less discussed in this course, but many of their properties may be familiar to you. A wave is a
movement of energy without any net motion of matter. (Water waves do not actually cause molecules of water to
move with the wave, only up and down.) Waves don't have a specific location in space, but are spread over some
area. This lack of localization is most strikingly demonstrated in the ability of waves to interfere, producing
interference or diffraction patterns.
Photons have properties of particles and waves simultaneously. This may seem contradictory, but the modern
quantum theory handles the contradictions perfectly. The real problem is when, for a particular situation, we try to
decide between treating photons approximately (without the full quantum theory) as waves or as particles. Usually
the case is clear cut.
For instance, photons of FM radio radiation should be treated as waves, since the energy of a single photon, about 108
eV, is much too small to produce a signal. Only the collective effect of billions of photons can be detected. Such a
large number of photons will not appear grainy, rather it will appear smooth, like a continuous, unquantized wave.
Interference and diffraction of radio waves is quite common.
On the other hand, an x-ray photon can have an energy of several thousand electronvolts, large enough for a single
photon to produce a substantial, detectable signal. X-ray photons should generally be treated individually, as
particles.
© Robert Harr 2000
course index
Recall from last lecture:

Photons behave sometimes like particles and sometimes like waves.
27.9 The Wave Properties of Particles
Having established that photons can have properties of both waves and particles, a physicist named Louis de Broglie
postulated that the same was true for matter. That is, he postulated that electrons, protons, neutrons, and even entire
atoms will exhibit wave-like properties under the proper circumstances.
For a photon, f = c, E = hf, and p = E/c = hf/c = h/. For matter, c = f is almost certainly not the case, and so how
do we determine the wavelength and frequency of a "matter wave"?
De Broglie suggested that the wavelength of a matter particle of momentum p = mv (non-relativistic particle) is
 = h / p = h / mv
and if the energy of the particle is E = mc⊃ (full relativistic energy), the frequency is
f = E / h.
Example: P27.39
Calculate the de Broglie wavelength of a proton moving at (a) 2.00×104m/s; (b) 2.00×108m/s.
(a) The momentum of the proton is
p = mv = (1.67×10-27kg)(2.00×104m/s) = 3.34×10-23kg m/s.
The de Broglie wavelength (the wavelength of the matter wave) is
 = h/p = (6.63×10-34Js)/(3.34×10-23kg m/s) = 1.99×10-11m.
(b) This proton is moving at at close to the speed of light. Let's calculate the momentum with the relativistic formula,
just to be safe:
p = mv = 4.48×10-19kg m/s
And then
 = h/p = (6.63×10-34Js)/( 4.48×10-19kg m/s) = 1.48×10-15m.
27.10 The Wave Function
The idea of particles behaving like waves caught on very quickly. Schrödinger put this idea to work in his wave
equation for quantum mechanics, the Schrödinger equation. Schrödinger's equation specifies how the wave function,
traditionally denoted by the greek letter psi, , behaves.
The wave function, , depends on location and time, and Schrödinger's equation tells us how to determine  for a
given situation. But, it doesn't tell us what  means. To understand what it means, let's look at what we know about
photons and electromagnetic waves.
Consider a double slit experiment with light. In the classical solution, we consider the light as a wave emerging from
each of the two slits in phase. At the viewing screen, we add up the electric fields of the waves from each slit, giving
us the resulting electric field. In some locations the electric fields add, giving bright fringes, and in others the electric
fields subtract, giving dark fringes.
Now consider this from the photon perspective. A single photon passing through the slits hits the screen at some
location. We can's predict ahead of time where it will hit, but we know that more photons must reach the area of a
bright fringe than the area of a dark fringe. We assign a probability to the photon hitting a particular location, and the
probability turns out to be proportional to the square of the electric field at that location, as predicted by the classical
solution, probability is proportional to E².
The electric field is the wave function of the photon. The square of the wave function at a particular location is
proportional to the probability of finding the photon at that location.
We extend this interpretation to the particle wave function: ² at some location is proportional to the probability of
finding the particle at that location (and time).
27.11 The Uncertainty Principle
The uncertainty principle is another well known modern Physics result. In words, the uncertainty principle says that it
is not possible to precisely know a particle's position and momentum (speed) at the same time. In a sense,
measurement of one disturbs the other.
Expressed as a formula, the uncertainty principle is
x px >= h/4
where x is the uncertainty in the particle's position, px is the uncertainty in the particle's momentum (in the same
direction), and the >= symbol means greater than or equal to. If the particle's position is known extremely accurately,
then its momentum cannot be well known. Conversely, if its momentum is accurately measured, then its position
cannot be well determined. The degree of uncertainty is regulated by the constant terms h/4.
A second uncertainty principle exists between energy and time:
E t >= h/4
This relation applies to, for instance, how well the energy of an excited state of an atom can be determined (by
measuring the width of its spectral line). The excited state will exist for a finite time, typically about 1ns, and then the
atom will decay to a lower energy state, and emit a photon of light. The energy of the excited state has a minimum
uncertainty given by E = h/4t = 3.3×10-7eV for a 1ns lifetime. The question of atomic line widths is important in
many laser applications.
Example: P27.46
In the ground state of hydrogen, the uncertainty in the position of the electron is roughly 0.10nm (the diameter of the
atom). If the speed of the electron is on the order of the uncertainty in the speed, how fast is the electron moving?
Using the uncertainty principle, we find
p = h / (4x) = (6.63×10-34Js)/4(1.0×10-10m) = 5.3×10-25kg m/s
Using the mass of the electron, we can find the uncertainty in the speed, which is approximately equal to the speed:
v = v = p/m = (5.3×10-25kg m/s) / (9.11×10-31kg) = 5.8×105m/s.
I used the non-relativistic expression for momentum, p=mv, and the fact that the resulting velocity is much less than
the speed of light is an a posteriori justification.
Example: P27.48
(a) Show that the kinetic energy of a non-relativistic particle can be written in terms of its momentum as KE = p²/2m.
(b) Use the result of (a) to find the minimum kinetic energy of a proton confined within a nucleus havin a diameter of
1.0×10-15m.
(a) Non-relativistic kinetic energy is KE = ½mv², and non-relativisitic momentum is p=mv. Therefore,
p²/2m = (mv)²/2m = ½mv² = KE.
(b) Since the proton is confined to the nucleus, x = 1.0×10-15m, and from the uncertainty principle,
p = h/4x = (6.63×10-34Js)/4(1.0×10-15m) = 5.3×10-20kg m/s.
The minimum average momentum of the proton is roughly equal to the uncertainty in the momentum, therefore, the
minimum kinetic energy is:
KEmin = p²min/2m = (p)²/2m = (5.3×10-20kg m/s)²/2(1.67×10-27kg) = 8.4×10-13J = 5.3×106eV = 5.3MeV.
© Robert Harr 2000
course index
Recall from last lecture:



Under the rules of quantum physics, the energy of a photon is E = hf.
Planck's constant, h = 6.63×10-34Js
Uncertainty principle: x px >= h/4
CHAPTER 28: ATOMIC PHYSICS
Throughout this chapter we will learn the Physics of the hydrogen atom. The hydrogen atom, with one proton and
one electron, is the most simple atomic system possible, and many results are derivable for this system. Some of these
results provide the most precise comparison between calculation and measurement ever achieved. Though derived for
this special case, the results we obtain are more general and tell us about the structure of all atoms, and the
organization of the periodic table of elements.
We begin with some history on the structure of atoms, then move to the approximate Bohr model and refinements.
The structure of the periodic table is discussed, and some applications will be discussed at the end.
28.1 Early Models of the Atom
"Atom" is a Greek word meaning the smallest piece, a whit. The original idea of an atom is traced to the Greek
philosopher Democritus who suggested that if you take something and cut or break it in half, again and again and
again, eventually you will be left with a piece that you could no longer break in two. This remaining piece is an atom.
In more modern usage, an atom is the smallest unit of an element that still retains the characteristics of that element.
We can't see atoms directly (or at least we couldn't until very recently), so we make a model that allows us to
rationalize how atoms behave. The earliest model, due to Newton, was a hard sphere, kind of like billiard balls.
Atoms will collide and elastically bounce off each other, conserving energy and momentum. This picture is used in
formulating the kinetic theory of gases, which generally works quite well.
When people realized that atoms are connected with the electrical properties of materials, then charge had to be
incorporated into the model of the atom. After J.J. Thompson demonstrated the existence of the electron, the "raisin
pudding" model came into vogue. This model consists of a sphere of uniform positive charge, with negatively
charged electrons stuck in it like raisins in pudding. The overall charge was zero, and electrons were incorporated, so
everyone was happy.
In 1911, Rutherford and his students performed an experiment, scattering alpha particles (He nucleus) off a thin gold
foil. They were very surprised when they observed some of the alpha particles scattering backward! An analogy for
this experiment is to imagine hitting a golf ball through a tree. If the golf ball is up near the small branches and
leaves, it will pass right through the tree, with only a small change from its original direction. If the golf ball hits the
trunk of the tree, it will bounce off at a large angle, maybe even coming right back at the golfer. This is what
Rutherford and his students observed. But the raisin pudding model of an atom doesn't have a "trunk", it doesn't have
a dense core for something to bounce off of. So Rutherford proposed a model with a small, dense core of positive
charge (the nucleus) surrounded by negatively charged electrons moving in orbits about the nucleus.
This model is closer to our present picture, but still has problems. First, it doesn't explain the spectral lines of atoms
(and molecules). Second, electrons in circular motion are accelerated, and, according to the equations of
electromagnetism, an accelerated charge will radiate electromagnetic energy, losing energy itself. The electrons
orbiting the nucleus should eventually radiate away all their kinetic energy, and fall into the nucleus.
Both these problems are solved by the application of quantum theory and Schrödinger's wave equation.
28.2 Atomic Spectra
You have seen spectral lines in one of the laboratory experiments. Different types of atoms emit different spectral
lines. In fact, the spectral lines can be used to fingerprint an atom. Even more common is to look at the absorption
lines produced, for instance, when light from a blackbody (a star) passes through a gas in space. The atoms in the gas
will selectively absorb certain frequencies, the same frequencies that are emitted when the atom is excited. In this
way, astronomers can determine what kind of atoms are in space, in what abundances, and even tell how fast the gas
is moving toward or away from us by the doppler shift of the lines.
The spectral lines of hydrogen are particularly interesting. Hydrogen emits in the visible at wavelengths of 656.3nm,
486.1nm, 434.1nm, and 410.2nm. In 1885, Johann Balmer realized that these wavelengths are described by the
formula:
1/ = RH{(1/2²) - (1/n²)}
where n is 3, 4, 5, ..., and RH = 1.0973732×10 m-1 is called the Rydberg constant.
7
Inserting n=3 into the Balmer formula gives the wavelength 656.3nm. Inserting n=4 gives 486.1nm, and so on.
Another set of spectral lines called the Lyman series is given by a similar formula:
1/ = RH{(1/1²) - (1/n²)}
The Lyman lines lie entirely in the ultraviolet part of the spectrum,  < 400nm.
The element helium was discovered because astronomers observed absorption lines from the sun that didn't
correspond to any known element. The name helium is derived from helios, greek for sun.
Example: P28.2
(a) Suppose the Rydberg constant in Balmer's formula were given by RH = 2.00×107m-1. What part of the
electromagnetic spectrum would the Balmer series correspond to? (b) Repeat for RH = 0.500×107m-1.
(a) The wavelengths in the Balmer series are given by
(1/) = RH{(1/2²) - (1/n²)}.
The shortest wavelength occurs for n = infinity, and the longest for n=3.
min = 2²/RH = 4 / 2.00×107m-1 = 2.00×10-7m = 200nm
max = 1 / (2.00×107m-1){1/4 - 1/9} = 3.6×10-7m = 360nm
Both of these wavelengths lie in the ultraviolet part of the spectrum, so the entire Balmer series lies in the ultraviolet.
(b) Repeat this with the other value for RH. We find:
min = 800nm
max = 1440nm
Now the Balmer series lies in the infrared.
28.3 The Bohr Theory of Hydrogen
Niels Bohr was a famous Danish physicist -- "Copenhagen", a recent play, is concerned with a conversation between
Bohr and Heisenberg (of the Heisenberg uncertainty principle) after World War II regarding the German program to
build a nuclear bomb. Bohr's model of the hydrogen atom is able to explain the spectral lines, and many other
features. The model is based on assumptions that mix together ideas from classical and quantum physics. The model
is superseded by Schrödinger's wave equation, but Bohr's model leaves us with a nice picture of an atom that is not
directly evident from the involved mathematics of the wave equation.
I quote the assumptions from the text:
1. The electron moves in circular orbits about the proton under the influence of the Coulomb force of attraction,
as in Figure 28.5. In this case, the Coulomb force is the force that produces centripetal acceleration.
2. Only certain electron orbits are stable. These are orbits in which the hydrogen atom does not emit energy in
the form of radiation. Hence, the total energy of the atom remains constant, and classical mechanics can be
used to describe the electron's motion.
3. Radiation is emitted by the hydrogen atom when the electron "jumps" from a more energetic initial state to a
lower state. The "jump" cannot be visualized or treated classically. In particular, the frequency, f, of the
radiation emitted in the jump is related to the change in the atom's energy and is independent of the frequency
of the electron's orbital motion. The frequency of the emitted radiation is
Ei - Ef = hf
where Ei is the energy of the initial state, Ef is the energy of the final state, h is Planck's constant, and Ei > Ef.
4. The size of the allowed electron orbits is determined by a condition imposed on the electron's orbital angular
momentum: The allowed orbits are thos for which the electron's orbital angular momentum about the nucleus
is an integral multiple of hbar = h/2.
me v r = n hbar
n = 1, 2, 3, ...
We can now use these assumptions to calculate the energy levels of the allowed orbits of the hydrogen atom.
© Robert Harr 2000
course index
Recall from last lecture:



Balmer series: 1/ = RH{(1/2²) - (1/n²)}, n=3, 4, ...
Lyman series: 1/ = RH{(1/1²) - (1/n²)}, n = 2, 3, ...
Rydberg constant, RH = 1.0973732×107m-1
28.3 The Bohr Theory of Hydrogen
I will repeat and paraphrase the four assumptions of the Bohr theory:
1. The electron moves in a circle about the proton under the influence of the Coulomb force.
2. Only certain orbits are allowed (meta-stable).
3. An electron can jump between allowed orbits by emitting or absorbing a photon with the proper energy given
by: Ei - Ef = hf.
4. The allowed orbits satisfy the angular momentum requirement:
mev r = n hbar
where n = 1, 2, 3, ..., and hbar = h/2.
From these assumptions we can solve for the allowed orbits and energy levels of the hydrogen atom. We begin with a
picture of the electron in a circular orbit around the proton (Figure 28.5). The electric potential energy of the atom is
PE = ke q1 q2 / r = -ke e2 / r
The total energy of the atom is the sum of the kinetic energy of the electron (the proton is at rest) and potential energy
E = KE + PE = (1/2)me v2 - ke e2 / r
(Notice that the hydrogen at is non-relativisitic, so we neglect the energy associated with the mass.)
For an object in circular orbit, the centripetal force must equal mv2/r. The centripetal force on the electron is the
Coulomb force:
ke e2 / r2 = me v2 / r
If we now multiply both sides by r and divide by 2, we see that:
ke e2 / 2r = (1/2) me v2
The left hand side is -1/2 the PE and the right hand side is the KE. Substituting this into the expression for the energy,
we get:
E = -ke e2 / 2r
From the condition for allowed orbits, we get:
vn = n hbar / me rn
where I've added the subscript n to remind us that this value depends on the energy level of the electron. If we square
this, we can compare to v2 from the centripetal force equation:
vn2 = n2 hbar2 / me2 rn2 = ke e2 / me rn
From the above expression, we can get an expression for the radii
rn = n2 hbar2 / me ke e2
where n can take on the integer values 1, 2, 3, ... The radius of the smallest orbit (n=1) is called the Bohr radius, a0
a0 = hbar2 / me ke e2 = 0.0529 nm
We can express the rn in terms of a0 as
rn = n2 a0
Now, substituting the expression for the radii into the energy expression we get:
En = -(me ke2 e4 / 2 hbar2) (1/n2)
This is our sought after result! If we substitute for the constants, we can write that the energy of the nth Bohr orbit is
En = -13.6eV / n2
The lowest energy state (n=1) has energy E1 = -13.6eV. The energy is negative because the electron is bound to the
proton, and we used the expression for potential energy which is zero when the electron and proton are widely
separated. This situation corresponds to n=infinity.
When an electron jumps between energy levels, the atom can emit or absorb a photon. Let's concentrate on emission,
for absorption just reverse the initial and final state. In emission, the initial state is just the atom in its initial energy
level, and the final state is the atom in its final energy level plus a photon. Conservation of energy give us:
Ei = Ef + hf
so that hf = Ei - Ef.
Supposing that the inital energy level is ni and the final level is nf, then the wavelength of the emitted photon is:
1/ = f/c = (Ei - Ef) / hc = (me ke2 e4 / 2hc hbar2) {(1/nf2) - (1/ni2)}
Replace h with 2 hbar, and the constant piece can be equated with Rydberg's constant:
RH = me ke2 e4 / 4 c hbar3
The agreement between the value of RH determined from measurements of the Balmer series, and this expression are
remarkably good. That agreement was recognized as a major accomplishment of Bohr's theory.
Draw energy levels for hydrogen and show transitions corresponding to Balmer and Lyman series (Figure 28.7).
Example: P28.9
Show that the speed of the electron in the nth Bohr orbit in hydrogen is given by vn = kee2 / n hbar.
From the last of Bohr's assumptions we have
mevn r = n hbar (1)
and equating the Coulomb force with the centripetal force gives
mevn2 / r = kee2 / r2. (2)
If we multiply (2) by r2 we get
mevn2 r = kee2. (3)
Now divide (3) by (1) and we get
vn = kee2 / n hbar.
Example: P28.12
Four possible transitions for a hydrogen atom are listed below:
(A) ni = 2; nf = 5 (B) ni = 5; nf = 3
(C) ni = 7; nf = 4 (D) ni = 4; nf = 7
(a) Which transition will emit the shortest wavelength photon? (b) For which transition will the atom gain the most
energy?
(a) First, to emit a photon, the initial level must be greater than the final level. This is only true for transitions (B) and
(C). Transitions (A) and (D) require absorption of a photon. Second, the shortest wavelength photon comes from the
transition with the largest energy difference. Using En = -(13.6eV / n2), and Ei = Ef + hf for emission, we find that for
transition (B)
Ei - Ef = -13.6eV (1/5² - 1/3²) = 0.967eV
and for transition (c)
Ei - Ef = -13.6eV (1/7² - 1/4²) = 0.572eV.
Therefore transition (B) will emit the shortest wavelength photon.
(b) The atom will gain energy when it absorbs a photon, Ei + hf = Ef. For transition (A), the change in energy is
Ef - Ei = -13.6eV (1/5² - 1/2²) = 2.86eV
and for transition (D)
Ef - Ei = -13.6eV (1/7² - 1/4²) = 0.572eV.
Therefore, the atom in transition (A) gains the most energy.
Example: P28.22 (optional)
Two hydrogen atoms, both initially in the ground state, undergo a head-on collision. If both atoms are to be excited to
the n=2 level in this collision, what is the minimum speed each atom can have before the collision?
Recall that collisions can be elastic (momentum and kinetic energy are conserved) or inelastic (momentum is
conserved, but not kinetic energy). In this problem, we want to convert some of the kinetic energy of the atoms into
internal energy (exciting the electrons to the n=2 level), therefore this is an inelastic collision. If kinetic energy is
converted to internal energy as efficiently as possible, then this is a totally inelastic collision -- the two atoms will be
stuck together afterwards. Momentum is conserved if the atoms have the equal and opposite initial momenta
(velocities), and, after colliding, are at rest.
Energy is conserved if the initial energy (kinetic energy plus ground state energy) equals the final energy (excited
state energy)
2E1 + 2K.E. = 2E2
Divide out the common factor of 2 such that
-13.6eV / 1² + (1/2)mHv2 = -13.6eV / 2²
where mH is the mass of a hydrogen atom, which is approximately the same as the mass of a proton, which is
938MeV / c². Solve for v
v2 = 13.6eV (1 - 1/4) 2 / mH = 20.4eV / 938MeV/c² = 2.2×10-8c²
v = 1.5×10-4c = 4.5×104m/s.
© Robert Harr 2000
course index
Recall from last lecture:
From Bohr's theory for hydrogen we have:




En = -(me ke2 e4 / 2 hbar2)(1/n2) = -13.6eV / n2
rn = n2a0
a0 = hbar2 / me ke e2 = 0.0529nm
RH = me ke2 e4 / 4 c hbar3
28.4 Modification of the Bohr Theory
Bohr's theory gives us a good model for the hydrogen atom, a model that is still used with a few modifications.
The Bohr theory can be applied to other "hydrogen-like" atoms (atoms with only one electron) by replacing the factor
e2 by Ze2 wherever it appears, where Z is the number of protons in the nucleus. For singly ionized helium, Z=2; for
doubly ionized lithium, Z=3; and so on.
The next refinement was to extend Bohr's theory to include elliptical orbits. This led to the concept of the orbital
quantum number, l. (The level n is known as the principal quantum number.) For a given energy level, n, the orbital
quantum number can take on values from 0 to n-1. The values of l are equated to the subshells: s is l=0, p is l=1, d is
l=2, and f is l=3.
The next refinement came when people noticed that what appeared to be a single spectral line would split into several
lines in the presence of a magnetic field (see Figure 28.9). This led to the concept of orbital magnetic quantum
number, ml. The results showed that ml could take on integer values from -l to +l.
Lastly, it was noted that at very high resolutions, single spectral lines split into two, even in the absence of a magnetic
field. This splitting is known as fine structure, and was connected to the concept of electron spin. The spin magnetic
quantum number, ms was introduced to account for this.
These four quantum numbers are what we still use to describe the structure of atoms: n, l, ml, and ms. We will discuss
these more in the next sections.
Example: P28.26
(a) Find the energy of the electron in the ground state of doubly ionized lithium, which has an atomic number Z=3.
(b) Find the radius of its ground-state orbit.
(a) To get the result for doubly ionized lithium, replace e2 by Ze2 in the expression for the energy levels.
En = (me ke2 Z2 e4 / 2 hbar2) (1/n2) = -Z2 (13.6eV / n2) = -122.4eV / n2
The ground state energy is
E1 = -122.4eV
(b) Make the same change to the expression for the radius:
rn = n2 hbar2 / me ke Z e2 = n2 a0 / Z
The ground-state radius is
r1 = a0 / Z = 0.0529nm / 3 = 0.0176nm.
28.5 De Broglie Waves and the Hydrogen Atom
Bohr's assumption that allowed orbits are those where the electron's orbital angular momentum is an integral multiple
of hbar results in correct predictions for the energy levels of hydrogen. But this assumption comes out of the blue.
Why it works was a mystery until de Broglie realized a connection between this assumption and his matter waves.
It turns out that Bohr's condition is equivalent to stating that the allowed orbits are those with an integral number of
oscillations of the electron's de Broglie waves. If an orbit has an integral number of oscillations, then the
circumfrence of the orbit is an integer multiple of the de Broglie wavelength, :
2 rn = n 
The de Broglie wavelength of an electron is  = h / me v. Therefore:
2 rn = n h / me vn
or, after rearranging terms,
me vn rn = n h / 2 = n hbar
This is Bohr's angular momentum requirement!
28.6 Quantum Mechanics and the Hydrogen Atom
The use of matter waves was expanded on by Schrödinger when he devised his wave equation for quantum
mechanics. One of the first problems analysed with the wave equation was the hydrogen atom. This problem can be
solved exactly, and some of the results are:



The energies of the allowed states agree exactly with the Bohr results.
The state of an atom is described by three quantum numbers,
o the principal quantum number, n
o the orbital quantum number, l
o the orbital magnetic quantum number, ml
The quantum numbers come directly out of the mathematics of the wave equation. It also determines the
allowed ranges of the quantum numbers:
o n ranges from 1 to infinity in integer steps
o l ranges from 0 to n-1 in integer steps
o ml ranges from -l to +l in integer steps
States that do not adhere to these rules are not allowed. The wave equations re-produces the results of the Bohr theory
plus modifications, without ad-hoc assumptions. Further, it can be applied to a host of other problems. The
significance is similar to the advent of Newton's laws, which not only re-produced the results of Kepler regarding the
orbits of the planets, but can then be used to solve many other problems.
28.7 The Spin Magnetic Quantum Number
There is a fourth quantum number that is needed to complete the atomic model. This fourth quantum number is ms,
the spin magnetic quantum number. We usually picture this quantum number as an electron spinning in one of two
possible directions (clockwise or counter-clockwise). This number was originally introduced to explain the splitting
of certain spectal lines, like the orange doublet of sodium.
Sodium is a group I element, like hydrogen, lithium, and potassium. We now understand that it consists of a complete
n=1 shell, a complete n=2 shell, plus one electron in the n=3 shell (1s22s22p63s1). This structure makes it "hydrogenlike". The ground-state energy depends on whether the outermost electron is "spinning" in the same direction as or
opposite to the nucleus of the atom. Thus, the ground-state energy "splits" into two closely spaced levels, and the
spectral line splits, one line corresponding to transitions to one of the levels and the other line to transitions to the
other level.
It is interesting to note that the spin of the electron and other particles is still put into the theory by hand. A
fundamental explanation of particle spin is still being sought.
Example: P28.32
When the principal quantum number is n=4, how many different values of (a) l and (b) ml are possible?
(a) The possible values of l range from 0 to n-1. For n=4, l can have 4 different values, 0, 1, 2, and 3.
(b) The possible values of ml range from -l to +l, irrespective of the value of n. Given the possible values of l from
part (a), we have the following possibilities for ml:
l ml
#
0 0
1
1 -1, 0, +1
3
2 -2, -1, 0, +1, +2
5
3 -3, -2, -1, 0, +1, +2, +3 7
Notice that the number of values of ml equals 2l + 1.
© Robert Harr 2000
course index
Recall from last lecture:



The wave theory for hydrogen predicts energy levels that agree exactly with the expressions found for the
Bohr theory.
The wave theory also says that the state of an electron is described by 3 quantum numbers, n, l, and ml.
The spin of the electron must be added to the theory by hand, and results in a fourth quantum number, ms.
28.8 Electron Clouds
The result of Schrödinger's wave equation for the hydrogen atom is a wave function, . There is a different wave
function for each set of quantum numbers. Recall that the square of the wave function tells us the probability of
finding the electron at a particular location.
If we look at the wave function for the n = 1, l = 0, ml = 0 state, we find that the wave function is resembles a
symmetric cloud (see Figure 28.13). That means it it equally probable to find the electron in any given direction from
the proton.
If the probability doesn't depend on angle, then it can only depend on the radius from the proton. When viewed in this
manner, we find that the probability peaks at a radius equal to the Bohr radius, a0 = 0.0529nm. It also happens that if
we look at the state with n = 2, l = 0, the probability peaks at a radius of 4a0, and for n = 3, at 9a0.
In the Bohr model, an electron follows a circular orbit at a fixed radius from the proton. The wave equation solution
indicates that the electron isn't in a circular orbit, however the radius of the Bohr orbit corresponds to the peak of the
probability distribution.
28.9 The Exclusion Principle and the Periodic Table
The state of an electron in an atom is described by the four quantum numbers n, l, ml, and ms. According to the
"exclusion principle", only one electron in an atom can have a given set of quantum numbers. The electrons in an
atom tend to move to the lowest available energy levels. Without the exclusion principle, they'd all end up in the n =
1 level.
Because of the exclusion principle, electrons fill up the available energy levels (shells) in order of increasing energy.
The lowest energy level is that with n = 1, l = 0, ml = 0, and ms = ±½. These are the levels occupied by the electron in
hydrogen and the two electrons in helium. We note their electron configurations as 1s1 for hydrogen and 1s2 for
helium. The first 1 indicates that the electrons are in the n = 1 level; the s indicates the l = 0 subshell; and the
superscript 1 or 2 indicates whether there are 1 or 2 electrons in this subshell.
This notation is readily extended to describe the electronic configuration of all atoms (see table 28.4). The subshells
with l = 0 are described with the letter "s", l = 1 with "p", l = 2 with "d", and l = 3 with "f". The "s" subshell can have
up to 2 electrons; the "p" subshell can have up to 6 electrons; the "d" subshell can have up to 10 electrons; and the "f"
subshell can have up to 14 electrons. These numbers are seen in the structure of the periodic table.
Example P28.38
Suppose two electrons in the same system each have l = 0 and n = 3. (a) How many states would be possible if the
exclusion principle were inoperative? (b) List the possible states, taking the exclusion principle into account.
(a) If the exclusion principle were inoperative, then either electron could be in any state consistent with l =0 and n =
3. The possibilities have ml = 0 (the only possible value) and ms = ±½. So we can arrange the two electrons in one of
three possible configurations: both in the ms = +½ state, both in the ms = -½ state, or one in each. The electrons are
still identical, so there's only one configuration with one in each, not two.
(b) Taking into account the exclusion principle, the electrons can't both be in the same state, so that leaves only one
possible configuration: one electron in ml = 0, ms = +½, and the other in ml = 0, ms = -½.
28.11 Characteristic X-Rays
As discussed in chapter 27, x-rays may be produced when an inner shell electron of a heavy atom is ejected, for
example by another electron striking the atom at high energy. When an electron in a high energy level jumps down to
fill the now vacant inner shell, an x-ray is emitted.
The emitted x-ray spectrum has a broad background with peaks corresponding to the characteristic x-rays of the
atom. We can estimate the energies of a characteristic x-ray using the technique outlined in the following example.
Example 28.7
Estimate the energy of the characteristic x-ray emitted from a tungsten target when an electron drops from an M shell
(n=3 state) to a vacancy in the K shell (n=1 state).
First estimate the energy of the K shell (n=1)level. This shell can hold two electrons, and since they are the innermost
electrons, it's as if they share the force due to all the protons in the nucleus. We can approximate this effect by saying
that for one of the electrons, the effective charge looks like the full charge of the nucleus minus one for the other K
shell electron, Zeff = Z - 1. Tungsten has Z = 74. Now treat this like a problem to calculate the ground state energy of
a hydrogen-like atom with Zeff protons:
EK = -Zeff2(13.6eV) = -(74 - 1)2 (13.6eV) = -72500eV.
The electron in the M shell (n=3) is shielded from the full nuclear charge by the electrons in the n=1 and n=2 shells.
The n=2 shell has 8 electrons (2 in s and 6 in p subshells), and the n=1 shell has 1 electron. Remember that one of the
n=1 electrons is missing, so that shell has only 1 electron, not 2. The effective charge seen by an M shell electron is
Zeff = Z - 9. To estimate the M shell energy, treat this like an n=3 level of a hydrogen-like atom with Zeff protons:
EM = -Zeff2(13.6eV)/32 = -(74 - 9)2(13.6eV)/32 = -6380eV.
The energy of the emitted x-ray is given by the difference between these levels:
Eph = EM - EK = -6380eV - (-72500eV) = 66100eV.
The wavelength of this x-ray is:
 = hc/Eph = (6.63×10-34Js)(3.0×108m/s)/(66100eV)(1.6×10-19J/eV) = 0.0188nm.
© Robert Harr 2000
course index
Recall from last lecture:
© Robert Harr 2000
course index
Recall from last lecture:
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Atomic number: Z = # protons in the nucleus
Neutron number: N = # neutrons in the nucleus
Mass number: A = # nucleons in the nucleus
r = r0 A1/3 where r0 = 1.2fm
29.1 Some Properties of Nuclei
Nuclear Stability
Nuclei are held together by the nuclear force, a force that is very strong over short distances -- strong enough to keep
the protons of a nucleus bound together despite their Coulomb repulsion.
Not all nuclei are stable. Stability means that a nucleus will remain as it is, essentially forever. Nuclei that are not
stable undergo nuclear decay. As a general rule, stable nuclei have as many, or more, neutrons as protons. And, the
more protons a nucleus has, the more neutrons needed for stability.
29.2 Binding Energy
The mas of a nucleus is always less than the sum of the masses of the protons and neutrons that make up the nucleus.
The difference in mass is due to the binding energy.
Ebind = (Zmp + Nmn - Mnucleus)c2
To get the binding energy per nucleon, divide by the mass number, A.
If we plot the binding energy per nucleon versus the number of nucleons, we see that there is a maximum binding
energy for mass numbers near 60 (Figure 29.4). These are the most stable nuclei.
Example: P29.8
Compare the average binding energy per nucleon of 2412Mg and 8537Rb.
In appendix B, we find the atomic masses:
M(2412Mg) = (23.985042)u and
M(8537Rb) = (84.911793)u.
From table 29.1 we find the mass of the proton and neutron are:
mp = (1.007276)u and
mn = (1.008665)u.
24
12Mg has 12 protons and 12 neutrons, so the binding energy is:
Ebind = 12(1.007276)u + 12(1.008665)u - (23.985042)u = 0.206u = 0.206(931 MeV/c²) = 192 MeV/c²
The number of nucleons is 24, so the binding energy per nucleon is:
Ebind / A = 8.00 MeV/c²
85
37Rb
has 37 protons and 48 neutrons, so:
Ebind = 37(1.007276)u + 48(1.008665)u - (84.911793)u = 0.773u = 0.773(931 MeV/c²) = 720 MeV/c²
The number of nucleons is 85, so the binding energy per nucleon is:
Ebind / A = 8.47 MeV/c²
29.3 Radioactivity
Radioactivity is the spontaneous emission of radiation by an unstable nucleus. There are three types of radiation
which are emitted by various nuclei:
1. alpha () rays (particles) which are 4He nuclei (2 protons and 2 neutrons),
2. beta () rays (particles) which are electrons, or, in a few cases, positrons (the anti-particle of the electron), and
3. gamma () rays which are high energy photons.
These three types of radiation can be distinguished by their behaviors. Gamma rays are not deflected (bent) by a
magnetic field; alpha particles are bent as for a positive charge, and most beta particles are bent as fro a negative
charge.
They are also distinguished by their penetrating power. Alpha particles are barely able to penetrate through a sheet of
paper. Beta particles can penetrate a few millimeters of aluminum or other light material. Gamma rays can penetrate a
few centimeters of lead or other heavy material.
The Decay Constant and Half-Life
Radioactive decay is a stochastic (random, probabilistic) process. It does seem to be a universal feature of radioactive
decay that the probability for a radioactive nuclei to decay during a period of time is a constant for that type of nuclei.
If we have N radioactive nuclei, then during time t, the number, N, that decay is given by
where  is the decay constant for that type of nuclei.
N =  N t
The activity, R, of a sample is the number of decays per second
R = N / t =  N
This leads to the result that, starting initially with N0 radioactive nuclei, the number, N, remaining (not having
decayed) a time t later is
N = N0 e-t
where e = 2.718... is the base of the natural logarithms. This is an exponential decay function.
The time it takes for half the initial nuclei to decay is called the half-life, T1/2, where the half-life is related to the
decay constant by:
T1/2 = ln(2) /  = 0.693 / 
There are 2 units for activity. The first is the curie (Ci), defined as 1Ci = 3.7×1010 decays/s. The second is the SI unit,
the becquerel (Bq), defined as 1Bq = 1 decay/s.
Example: P29.14
A drug tagged with 4399Tc (half-life = 6.05h) is prepared for a patient. If the original activity of the sample was
1.1×104Bq, what is its activity after it has sat on the shelf for 2.0h?
The activity is proportional to the number of undecayed nuclei in the sample, R = N. If the inital activity is R0 = N0
= 1.1×104Bq, then after time t, the activity will be
R = N = N0 e-t = R0e-t.
We can determine  from the half-life:
 = 0.693 / T1/2 = 0.693 / 6.05h = 0.1145/h.
Therefore:
R = (1.1×104Bq)e-[(0.1145/h)(2.0h)] = 0.87×104Bq = 8.7×103Bq.
© Robert Harr 2000
course index
Announcements:
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The problems from chapter 30 can be omitted. They will not be used on the final exam.
For the final exam, you are expected to know the values of the metric prefixes from pico (10-12) to Giga (109).
These will not be given on the equation sheet.
Recall from last lecture:
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Nuclei are composed of neutrons and protons.
A nucleus of element X with Z protons, N neutrons, and A = Z + N neutrons is indicated as AZX.
Ebind = (Z mp + N Mn - Mnucleus)c2.
29.4 The Decay Process
When a nucleus decays, it is transformed in some way. This transformation is called the decay process, and is
indicated by a decay formula such as X --> Y + Z. As in chemistry, there are rules for valid formulas:
1. Charge is conserved in a decay process.
2. The number of nucleons (the mass number) is conserved.
Let's look at how these apply to each of the three decay types.
Alpha Decay
In alpha decay the original nucleus (called the parent nucleus) breaks up into a decay nucleus (called the daughter)
and a 42He nucleus. Since charge and the number of nucleons are both conserved, if the parent nucleus is AZX, then
the daughter nucleus is A-4Z-2Y, where X is the element with atomic number Z, and Y is the element with atomic
number Z-2.
Beta Decay
In beta decay, a neutron in the parent nucleus convertes into a proton, an electron, and a third particle called an antineutrino. The anti-neutrino, written as the greek letter nu with a bar over top (nubar), is neutral, massless, and very
hard to detect, but its presence is evident from the range of electron energy in the decay. Here charge is conserved,
since the proton and electron charge in the final state ad up to zero. In beta decay, the daughter nucleus has the same
mass number, A, as the parent, but the atomic number increases by 1. An example beta decay formula is written:
A
A
ZX --> Z+1X + e + nubar
Gamma Decay
Gamma decay occurs when a nucleus in an excited state decays to a lower state, and emits a gamma ray. In gamma
decay the number of protons, and neutrons is unchanged.
Example: P29.24
Complete the following radioactive decay formulas:
12
5B --> ? + e + nubar
234
230
90Th -->
88Ra + ?
14
? --> 7N + e + nubar
The first decay is clearly a beta decay. A neutron in the parent nucleus is transformed into a proton (which remains in
the daughter nucleus), an electron, and an anti-neutrino. Therefore, the daughter nucleus has the same number of
nucleons as the parent, but one less neutron and one more proton, for a total of 6 protons. Therefore, the daughter
nucleus is carbon:
12
12
5B --> 6C + e + nubar
The second decay is an alpha decay since the resulting nucleus has 4 fewer nucleons and 2 fewer protons than the
parent nucleus. This decay is:
234
230
4
90Th -->
88Ra + 2He
The third decay is a beta decay. The parent nucleus will have one less proton and one additional neutron than the
daughter nucleus. Therefore the parent nucleus has 6 protons and is carbon:
14
14
6C --> 7N + e + nubar
29.5 Natural Radioactivity
omit from lecture
29.6 Nuclear Reactions
Nuclear reactions occur when two nuclei collide. Rutherford first did this in the laboratory in a variation of the
famous scattering experiment. He collided alpha particles with nitrogen nuclei, observing an outgoing hydrogen:
4
+ 147N --> X + 11H
To determine the nature of X, we balance the neutrons and protons on both sides. The left side of the reaction has 9
protons and 9 neutrons. The right side must have as many, therefore X must have 8 protons and 9 neutrons, since
hydrogen has but 1 proton. The element with 8 protons is oxygen, therefore X is 178O.
© Robert Harr 2000
2He
course index
Recall from last lecture:
CH. 30: ELEMENTARY PARTICLES
We once thought that atoms were the basic building blocks of the universe. Then, we discovered that atoms are
composed of protons, neutrons, and electrons. For several decades we thought that these were the basic building
blocks, but starting in the 1960's we discovered that protons and neutrons are composed of even smaller particles
called "quarks".
The field of physics dedicated to understanding the fundamental constituents of the universe is called "elementary
particle" physics. This field emerged from nuclear physics, and nowadays shares common interests with nuclear
physics, astrophysics, and even atomic physics.
The Fundamental Particles
All known particles fall into three categories: quarks, leptons, and gauge bosons. The quarks and leptons are of a class
called fermions, and are the basic constituents of all matter. The gauge bosons are the carriers of force.
Quarks
The proton and neutron are composed of quarks, specifically, "up" and "down" quarks. The distinguishing feature of
quarks is that they "feel" the strong force, the force which binds the up and down quarks into protons and neutrons,
and is discussed later. In addition to up and down quarks, we have discovered four other quarks, known as "strange",
"charm", "beauty" or "bottom", and "top". We normally indicate the quarks with the lower case letter at the start of
the name, so we have d, u, s, c, b, t. (The b quark is often called bottom, because it is naturally paired with the top
quark. However, people studying processes involving the b quark prefer to refer to it as "beauty physics"; the term
"bottom physics" too closely resembles the deragatory term "bottom feeders".)
The six quarks separate into two groups with different electric charge: u, c, and t have charge +(2/3)e, and d, s, and b
have charge -(1/3)e. There is also a natural progression in mass. The u and d quark are relatively light, the c and s
quarks are intermediate in mass, and the b and t quarks are heavy. The t quark is the heaviest of all. The mass of a
single t quark is roughly equivalent to the mass of an atom of gold, 175GeV. Due to its large mass, the t quark was
the last to be discovered. It was first observed in 1995 at Fermilab, a major national research laboratory about 40
miles from Chicago.
Leptons
Electrons fall into the lepton category. Leptons are fermions that don't feel the strong force. The other leptons are the
muon, the tau, and three "neutrinos", the electron neutrino mentioned in beta decay, the muon neutrino, and the tau
neutrino.
The neutrinos have zero charge, and zero or extremely small mass. This combination makes them very hard to detect,
they don't interact electrically, and their weak interactions are, well, weak. Yet they are so abundant, that if they have
even the miniscule mass of a fraction of an eV (about one-millionth the mass of an electron), their mass would
roughly equal the mass of all the atoms in the universe! If we want to understand the universe, then we must
understand neutrinos.
The electron, muon, and tau are charged leptons with charge -1e. These particles are basically identical except for
their masses. The muon and tau are sometimes referred to as heavy copies of the electron. Why nature has two
additional copies of the electron is still a mystery.
Gauge Bosons
In our present view of the world, force is due to the exchange of a particle from the category of "gauge bosons". To
every force there corresponds a gauge boson, and to every gauge boson a force. The idea is something like two ice
skaters tossing a mass between themselves. When skater 1 throws the mass, she is pushed in the opposite direction so
that momentum is conserved (momentum of skater 1 = -momentum of mass). When skater 2 catches the mass, he is
also pushed, this time in the direction of the moving mass. In the case of the exchange of gauge bosons, the idea is
similar except that, through the magic of Heisenberg's uncertainty principle, the force can be attractive or repulsive!
We have already encountered one of the gauge bosons, the photon. The photon is the "carrier" of the electromagnetic
force -- both the electric and magnetic force. The photon is massless, and susceptible only to gravity.
Having mentioned gravity, it's gauge boson is called the graviton. The graviton is extremely interesting, but it's most
important role is in astrophysics, and string theory. I won't discuss it further.
The next gauge boson goes by the name "gluon". The gluon is the carrier of the strong force which binds the quarks
into protons, neutrons, and similar particles. The name comes from the picture of binding the quarks with "glue",
making it basically impossible to have a single free quark. While there is basically one photon and one graviton, there
are 8 different gluons.
The final gauge bosons are a triplet known as the weak bosons because they carry the "weak" force responsible for
beta decay. These bosons are known as W+, W-, and Z0. Many of the modern advances in particle physics are due to
studying these 3 bosons.
The Fundamental Forces
The modern picture of forces is of an exchange of a force carrying particle between two other particles. We recognize
4 fundamental forces in the universe.
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The strong force which binds quarks
The Standard Model of Particle Physics
© Robert Harr 2000