Download Quantum Mechanics Unit Review Answers AP Physics

Quantum Mechanics Unit Review Answers
AP Physics
Conceptual Questions:
1. How does the idea that light energy is quantized explain the ultraviolet catastrophe (the fact that less
light energy is put out at high frequencies by a blackbody than would be expected based on early
theory).
The idea that light energy is quantized explains the lack of high frequency light put off by a
blackbody because high frequency light would come in high energy quanta. To produce a single
quantum (now called a photon) of high frequency light requires a large amount of energy (on the
atomic scale) and there is a limit to how much energy any individual electron or proton in a
blackbody possesses to create a photon. As you look at higher and higher frequency light, there are
fewer and fewer particles capable of creating that quantum and so fewer of those photons are made
until you get to high enough frequency that none of that quantum can be made by that object.
If you raised the object’s temperature that would give more energy to be divided among the
particles inside the object, which would raise the maximum amount of energy any individual particle
could have and therefore, the energy of the photon it could create and the frequency of the light it
puts off
2. How many photons are absorbed by any one electron in a photoelectric material? How do(es) the
absorbed photon(s)’s energy affect the electron?
An electron absorbs one photon at any time. Absorbing a photon gives energy to the electron.
This goes into raising the electron’s potential energy (moving it further away from the nucleus) and
raising the electron’s kinetic energy. If the photon has enough energy, it will first raise the potential
energy enough that the electron is ionized or separated from the nucleus (the amount of energy
required to do this is called the work function) and any energy left over after that goes into kinetic
energy.
3. What happens to the wavelength of a photon that undergoes Compton scattering off a free electron?
Why?
The wavelength of a photon that is scattered off an electron will increase.
When a photon hits a free electron, the electron rebounds from the impact. The electron cannot
simply absorb all of the photon’s energy because then momentum would not be conserved (an
electron and a photon will not have the same amount of momentum if they have the same amount of
energy). This means that the photon must “rebound” off of the electron and come out of the
collision traveling in some direction.
As the electron flies away from the collision, it has kinetic energy that must have come from the
photon that hit it. This means that the photon that comes out of the collision has less energy than it
did before the collision. For a photon, lower energy means lower frequency and lower frequency
implies a longer wavelength.
4. a) How does Bohr’s model of the hydrogen atom explain why gaseous elements that are stimulated
emit line spectra?
Bohr’s model of the hydrogen atom says that the energy of the electrons in orbit around the
nucleus is quantized – only particular values are allowed. The reason that gaseous elements only
release particular lines is that the electron can only give up the specific amount of energy between
one orbit and another. Because the energy released as electrons change orbits is emitted as a photon,
only photons with energy matching the gap between orbits will ever be emitted – those photons, with
their specific frequency and wavelength are the ones that make one particular line. Because there
are many different orbits and many more different combinations of starting orbit and finishing orbit,
there will be many different energy photons released from the gas, each one creating a different line.
b) Why do those elements absorb the same wavelengths from a continuous spectrum passing through
them that they would emit on their own?
When an atom absorbs a photon, the electron jumps from a lower orbit to a higher orbit. The only
photons that can be absorbed are the ones that have exactly the right amount of energy to lift the
electron to a particular higher orbit. That amount of energy is the energy of the photon that is
emitted as the electron falls back down to the lower level where it started. Therefore the lines in the
emission spectrum and absorption spectrum have the same frequency and wavelength.
5. What components are necessary to build a laser?
A laser must have a material that will “lase”, that is which will undergo stimulated emission. For
this to happen, the material should have a metastable high-energy state where electrons can remain
for some time. This allows a population inversion to develop and be maintained.
To get the electrons up to the metastable high-energy state without stimulating emission, it is
useful if the material will also fluoresce, so you can lift it up to a higher energy state from which it
will quick decay into the metastable state.
The final piece is a set of good (but not perfect) mirrors to keep the photons bouncing back and
forth, continually stimulating emission. At least one of the mirrors must let some of the photons out
to create the laser beam.
6. Why do we believe DeBroglie’s idea that particles like an electron are also waves in some way?
There are two major points that makes DeBroglie’s idea convincing: electron diffraction and
Bohr’s orbits.
It turns out that electrons shot through two tiny slits will produce an interference pattern on the
screen behind the slits, exactly like light with the same wavelength. Realistically, you can’t
construct slits small enough to diffract electrons, but you can use the regular spacing between atoms
in a crystalline solid as a set of slits that serves admirably, producing exactly the results that the
wave theory of electrons predicts.
The explanation of why Bohr’s model only includes the particular orbits it does was perhaps more
significant when the theory was first proposed. If electrons are waves, they can only exist in orbits
where the wave reinforces itself by ending up in the same phase at the end of the orbit as it did in the
beginning of the orbit so that the ends of the wave constructively interfere. This means that the only
orbits allowed in Bohr’s model are the ones where an integer number of electron waves fit around
the circumference of the orbit.
7. Why is the quantum mechanical model of the atom different from Bohr’s model of the hydrogen atom?
Because DeBroglie had already shown that Bohr’s model fits with a wave model of electron behavior,
there was no need to adjust the model simply to include that revolutionary idea. However, the quantum
mechanical model of the atom does make two major changes to Bohr’s model.
First it does away with Bohr’s orbits because the Heisenberg uncertainty principle won’t allow enough
precision in an electron’s position and momentum to maintain a nice circular orbit. Orbits are replaced
with orbitals which are regions of space where the electron is likely to be found.
Second, the orbitals occupy all three dimensions of space. This opens up more variety in orbital type.
Bohr’s orbits only varied in size but quantum mechanical orbitals vary in size, shape and complexity.
They have three quantized numbers that describe them instead of just one: n, the principal quantum
number, describes how the wave fits in the radial direction (away from the nucleus) and sets the base
level for the energy of the orbital; l, the angular momentum quantum number describes how the wave fits
around the circumference of the orbital [incidentally telling us how much angular momentum an electron
in that orbital will have] and affects the way an electron in that orbital will respond to other electrons
[which is expressed in the electron’s energy in that orbital]; and ml, the magnetic quantum number,
describes the orientation of an orbital with angular momentum, which affects the energy of the orbital
when an external magnetic field is applied to the atom (called the Zeeman effect).
[There is one other quantum number that applies to electrons in an atom, but it a property of the
electron itself, not the orbital. The electron’s spin, ms, tells the way the electron’s intrinsic angular
momentum (separate from orbital angular momentum) is pointing and the way a magnetic field will
affect the electron]
8. The same types of orbits exist in every atom that there is, though they become more appealing to an
electron in an atom that has more protons.
a) How does having more electrons change the energy of orbits in an atom?
Electrons repel each other, which results in a higher potential energy for the electrons in multielectron atoms. Because the potential energy of an electron in an atom is negative, this means that
other electrons makes the potential energy of the electron less negative, bringing it closer to 0 J and
freedom.
All orbits end up at a higher energy because of the repulsion between electrons, but some orbits
experience the repulsion to a greater degree. The higher the angular momentum quantum number of
the orbit that the electron is in – this is the number we represent with l, which describes the wave’s
circumferential behavior – the more the electron’s energy will be raised.
In an atom with only one
electron, all orbits with the same
principle quantum number – this is
the number we represent with n,
which describes the wave’s radial
behavior – have the same energy.
In an atom with more than one
electron, the orbits in the same
energy level with have different
energies, based on their angular
momentum quantum numbers.
The l = 1 orbits will be at slightly
higher energy than the l = 0 orbits
for the same shell, as shown in the picture.
b) How does an external magnetic field change the energy of the orbits in an atom?
An external magnetic field will
adjust the energy of the orbits in
different ways depending on how the
orbit is oriented – which is indicated
by the magnetic quantum number, m.
The orbits with higher m values will
be adjusted in energy by a greater
amount, producing an energy level
diagram like the one shown to the
left. Notice that this means we will
perceive an energy difference in the
photon given off when an electron
drops from or into different orbits with the same l value but different m values. This shows up on
the object’s spectrum as multiple lines produced where there would only be one if there were no
magnetic field. This
picture shows how one
spectrum might look
different in a magnetic
field.
Practice Calculations:
1. When a monochromatic light with a wavelength of 254 nm falls onto the surface of a particular metal
it causes a photocurrent to flow. A stopping voltage of 2.30 V is required to totally block the
photocurrent.
a) What is the work function of the material?
Ephoton = hf = Kmax + 0
0 = hf - Kmax
and
Kmax = eV
8
3 x 10 m/s
hf = 4.136 x 10-15 eV·s x 254 x 10-9 m = 4.89 eV and Kmax = e x 2.30 V = 2.30 eV
0 = hf - Kmax = 4.89 eV – 2.30 eV = 2.59 eV
b) What is the cutoff wavelength for this metal?
Cutoff value is when photon has just enough energy to overcome work function and can give no
kinetic energy to the electron.
hf0 = 0
0
2.59 eV
f0 = h = 4.136 x 10-15 eV·s = 6.262 x 1014 Hz
c
3 x 108 m/s
 = f = 6.626 x 1014 Hz = 4.79 x 10-7 m = 479 nm
0
c) Will light with a wavelength of 523 nm be able to cause a photocurrent from this material?
No, it has a longer wavelength and therefore a lower frequency than 479 nm light, which means
that the energy of one photon is lower than the photon energy for 479 nm light. Lower energy
means that it will not be able to add enough energy to “overcome” the work function and free an
electron.
2. A different metal has a cutoff wavelength of 365 nm.
a) What is the work function of this metal?
c
hf0 = 0 and
=f
c
3 x 108 m/s
f0 =
= 365 x 10-9 m = 8.219 x 1014 Hz
0
0 = hf0 = 4.136 x 10-15 eV·s x 8.219 x 1014 Hz = 3.40 eV = 5.44 x 10-19 J
b) What stopping voltage would be required to prevent a monochromatic light of 67.4 nm from creating
a photocurrent?
hf = Kmax + 0
and
eV = Kmax
Kmax
Kmax = hf - 0
and
V= e
c
3 x 108 m/s
f = = 67.4 x 10-9 m = 4.451 x 1015 Hz

Kmax = hf - 0 = 4.136 x 10-15 eV·s x 4.451 x 1015 Hz – 3.40 eV = 15.01 eV
Kmax 15.01 eV
V= e =
= 15.01 V
e
3. An experiment involves scattering 0.071 nm photons off of free electrons. What is the wavelength of
the scattered photons at each of the following angles?
a) 30°
b) 70 °
c) 130°
h
6.626 x 10-34 J·s
a)  = m c(1 – cos) = 9.11 x 10-31 kg x 3 x 108 m/s(1 – cos30) = 3.24 x 10-13 m = 0.000324 nm
0
f = i +  = 0.071 nm + 0.000324 nm = .071324 nm
h
6.626 x 10-34 J·s
b)  = m c(1 – cos) = 9.11 x 10-31 kg x 3 x 108 m/s(1 – cos70) = 1.59 x 10-12 m = 0.00159 nm
0
f = i +  = 0.071 nm + 0.00159 nm = .07259 nm
h
6.626 x 10-34 J·s
c)  = m c(1 – cos) = 9.11 x 10-31 kg x 3 x 108 m/s(1 – cos130) = 3.98 x 10-12 m = 0.00398 nm
0
f = i +  = 0.071 nm + 0.00398 nm = .07498 nm
4. Using the Bohr model of the hydrogen atom, calculate the orbital radius, orbital speed and energy
for the following energy levels:
a) n = 2
b) n = 3
c) n = 6
d) n = 7
n2h2
= 0.05298 nm x n2
4mke2
nh
n
vn =
= 1.164 x 10-4 m2/s r
2mr
2mk2e4 1
1
En = - ( h2 )n2 = -13.6 eV x n2
rn =
a) r2 = 0.05298 nm x 22 = 0.21192 nm
2
v2 = = 1.164 x 10-4 m2/s
= 1.0985 x 106 m/s
0.21192 x 10-9 m
1
E2 = -13.6 eV x 22 = -3.4 eV
b) r3 = 0.05298 nm x 32 = 0.47682 nm
3
v3 = = 1.164 x 10-4 m2/s 0.47682 x 10-9 m = 7.324 x 105 m/s
1
E3 = -13.6 eV x 22 = -1.51 eV
c) r6 = 0.05298 nm x 62 = 1.9073 nm
6
v6 = = 1.164 x 10-4 m2/s 1.9073 x 10-9 m = 3.662 x 105 m/s
1
E6 = -13.6 eV x 62 = -0.38 eV
d) r7 = 0.05298 nm x 72 = 2.5960 nm
7
v7 = = 1.164 x 10-4 m2/s 2.5960 x 10-9 m = 3.129 x 105 m/s
1
E7 = -13.6 eV x 72 = -0.28 eV
5. Find the photon energy and wavelength of the light emitted for each of the following quantum leaps
inside a hydrogen atom: (note – check out problem 6)
a) n = 6 to n = 2
b) n = 7 to n = 2
c) n = 7 to n = 6
Ephoton = Ei - Ef
f=
E
h
=
c
f
a) Ephoton = E6 – E2 = -0.38 eV – (-3.4 eV) = 3.02 eV = 4.83 x 10-19 J
E
4.83 x 10-19 J
f = h = 6.626 x 10-34 J·s = 7.30 x 1014 Hz
c
3 x 108 m/s
 = f = 7.30 x 1014 Hz = 4.10 x 10-7 m = 410 nm
b) Ephoton = E7 – E2 = -0.28 eV – (-3.4 eV) = 3.12 eV = 4.99 x 10-19 J
E
4.99 x 10-19 J
f = h = 6.626 x 10-34 J·s = 7.54 x 1014 Hz
c
3 x 108 m/s
= =
= 3.98 x 10-7 m = 398 nm
f 7.54 x 1014 Hz
c) Ephoton = E7 – E6 = -0.28 eV – (-0.38 eV) = 0.1 eV = 1.602 x 10-20 J
E
1.602 x 10-20 J
f = h = 6.626 x 10-34 J·s = 2.42 x 1013 Hz
c
3 x 108 m/s
 = f = 2.42 x 1013 Hz = 1.24 x 10-5 m = 12.4 m
6. Calculate the deBroglie wavelength for an electron in the second and third energy levels of a Bohr
atom. Check that an integer number of wavelengths fit around the circumference of the orbit.
h h
 = p = mv
and based on the results from question 5
h h
a)  = p = mv
h h
b)  = p = mv
6.626 x 10-34 J·s
= (9.11 x 10-31 kg)(1.0985 x 106 m/s)
6.626 x 10-34 J·s
= (9.11 x 10-31 kg)( 7.324 x 105)
= 6.62 x 10-10 m = 0.662 nm
= 9.931 x 10-10 m = 0.9931 nm
C2 = 2r2 = 2(0.21192 nm) = 1.33 nm
C2 = 2r3 = 2(0.47682 nm) = 2.996 nm
C 1.33 nm
=
= 2.009 ≈ 2
 0.662 nm
C 2.996 nm
=
= 3.017 ≈ 3
 0.9931 nm
7. a) You shine a red light on an electron and measure its position with an uncertainty of 750 nm.
What is the minimum uncertainty about its momentum? How much uncertainty in its speed?
h
h
6.626 x 10-34 J·s
x·p ≥
so
p ≥
=
= 1.41 x 10-28 kg·m/s
2
2·x 2·750 x 10-9 m
b) You switch to a blue light and measure the position with an uncertainty of 450 nm. What is the
minimum uncertainty in its momentum? How much uncertainty in its speed?
h
h
6.626 x 10-34 J·s
x·p ≥
so
p ≥
=
= 2.34 x 10-28 kg·m/s
2
2·x 2·450 x 10-9 m
8. a) How many different emission lines can be produced by an electron jumping from the third energy
level of a single-electron atom to the 2nd energy level of an atom under normal conditions?
There are 9 different orbitals in the 3rd energy level, with quantum numbers (n, l, ml): (3,0,0),
(3,1,1), (3,1,0), (3,1,-1), (3,2,2), (3,2,1), (3,2,0), (3,2,-1), (3,2,-2). In a single-electron atom, all
orbitals in the same energy level are degenerate – an electron in them has exactly the same amount
of energy. So there is only one energy value possible in the 3rd energy level in a single-electron
atom.
There are 4 different orbitals in the 2nd energy level, with quantum numbers (n, l, ml): (2,0,0),
(2,1,1), (2,1,0), (2,1,-1). In a single-electron atom, all orbitals in the same energy level are
degenerate – an electron in them has exactly the same amount of energy. So there is only one energy
value possible in the 2nd energy level in a single-electron atom.
This means that an electron jumping from the 3rd energy level to the 2nd energy level can only lose
1 amount of energy no matter which orbital they jump from and which they jump to so there will
only be one photon energy and one emission line produced by electrons jumping between those
levels
b) How many different emission lines can be produced by an electron jumping from the third energy
level of a multi-electron atom to the 2nd energy level of an atom under normal conditions?
There are 9 different orbitals in the 3rd energy level, with quantum numbers (n, l, ml): (3,0,0),
(3,1,1), (3,1,0), (3,1,-1), (3,2,2), (3,2,1), (3,2,0), (3,2,-1), (3,2,-2). In a multi-electron atom, there is a
difference in energy for orbitals with different values of angular momentum (l), though orbitals with
different values of the magnetic quantum number are degenerate. So there are three different energy
value possible in the 3rd energy level in a single-electron atom.
There are 4 different orbitals in the 2nd energy level, with quantum numbers (n, l, ml): (2,0,0),
(2,1,1), (2,1,0), (2,1,-1). In a multi-electron atom, there is a difference in energy for orbitals with
different values of angular momentum (l), though orbitals with different values of the magnetic
quantum number are degenerate. So there is are two energy values possible in the 2nd energy level in
a single-electron atom.
This means that an electron jumping from the 3rd energy level to the 2nd energy level can lose 6
different amounts of energy depending on which l-value orbital they jump from and which they
jump to so there will be six photon energies and six emission lines produced by electrons jumping
between those levels
c) How many different emission lines can be produced by an electron jumping from the third energy
level of a multi-electron atom to the 2nd energy level of an atom in a strong magnetic field?
There are 9 different orbitals in the 3rd energy level, with quantum numbers (n, l, ml): (3,0,0),
(3,1,1), (3,1,0), (3,1,-1), (3,2,2), (3,2,1), (3,2,0), (3,2,-1), (3,2,-2). In a multi-electron atom in a
magnetic field, there is a difference in energy for all different orbitals. So there are nine different
energy values possible in the 3rd energy level in a single-electron atom.
There are 4 different orbitals in the 2nd energy level, with quantum numbers (n, l, ml): (2,0,0),
(2,1,1), (2,1,0), (2,1,-1). In a multi-electron atom, there is a difference in energy for orbitals with
different values of angular momentum (l), though orbitals with different values of the magnetic
quantum number are degenerate. So there is are four energy values possible in the 2nd energy level
in a single-electron atom.
This means that an electron jumping from the 3rd energy level to the 2nd energy level can lose 36
different amounts of energy depending on which l-value and ml-value orbital they jump from and
which orbital they jump to so there will be 36 photon energies and 36 emission lines produced by
electrons jumping between those levels