Download Quantum Mechanics Unit Review AP Physics

Quantum Mechanics Unit Review
AP Physics
Conceptual Questions:
1. How does the spectrum of light emitted by a blackbody change as the blackbody increases in
temperature?
The intensity increases at all wavelengths and the maximum intensity moves to a lower wavelength.
2. 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 them are made until you get
to high enough frequency that none of that quantum can be made by that object.
3. 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.
4. What happens to the wavelength of a photon that undergoes Compton scattering off an electron?
Why?
The wavelength of a photon that is scattered off an electron will increase. This happens because the scattered photon
loses some energy to scatter the electron. For a photon, lower energy means lower frequency and lower frequency
implies a lower wavelength.
5. 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 the particular lines.
b) Why do those elements absorb the same wavelengths from a continuous spectrum passing through
them that they would emit on their own?
When elements absorb 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 in 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.
6. 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 must 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, the material must also fluoresce, so you can lift it up to a higher energy state from which it will 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.
7. Why is the quantum mechanical model of the atom different from Bohr’s model of the hydrogen atom?
The quantum mechanical model of the atom makes two major changes to Bohr’s model.
First it does away with Bohr’s orbits because the 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, sets the base level for the energy of
the orbital; l, the angular momentum quantum number, tells the angular momentum of the orbital, which affects the
energy of the orbital in multi-electron atoms; 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 (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]
Practice Calculations:
1. What is the maximum intensity wavelength for a blackbody at each of the following temperatures?
Identify the wavelength and what region of the spectrum it falls into:
a) 98.6 °F (Human body)
b) 20 °C (room temp.)
c) 78 K (LN2)
2.898 x 10-3 m 2.898 x 10-3 m
a) max =
=
= 9.34 x 10-6 m
infrared
T
310 K
2.898 x 10-3 m 2.898 x 10-3 m
b) max =
=
= 9.89 x 10-6 m
infrared
T
293 K
2.898 x 10-3 m 2.898 x 10-3 m
c) max =
=
= 3.72 x 10-5 m
infrared
T
78 K
2. a) The maximum intensity of light produced by the sun falls at a wavelength of 501 nm. What is the
surface temperature of the sun? (Assume the sun is a blackbody. It is quite close.)
2.898 x 10-3 m 2.898 x 10-3 m
T=
= 501 z 10-9 m = 5784 K
max
b) An average value for the maximum intensity wavelength of light emitted from the earth is 9.89 x 10-6
m. (We use an average value because there is a difference between the daylight side of the earth and the
night side of the earth, not to mention the poles and the equator…) What is the average surface
temperature of the earth? (Assume the earth is a blackbody. It is quite close.)
2.898 x 10-3 m 2.898 x 10-3 m
T=
=
= 0.0034 K
0.835 m
max
corrected:
2.898 x 10-3 m 2.898 x 10-3 m
T=
=
= 293 K
9.89 x 10-6 m
max
3. 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.
4. 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
5. An experiment involves scattering 0.071 nm photon 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
6. 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
= 0.059 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 = 0.059 nm x 22 = 0.236 nm
2
v2 = = 1.164 x 10-4 m2/s 0.236 x 10-9 m = 9.86 x 105 m/s
1
E2 = -13.6 eV x 22 = -3.4 eV
b) r3 = 0.059 nm x 32 = 0.531 nm
3
v3 = = 1.164 x 10-4 m2/s 0.531 x 10-9 m = 6.57 x 105 m/s
1
E3 = -13.6 eV x 22 = -1.51 eV
c) r6 = 0.059 nm x 62 = 2.124 nm
6
v6 = = 1.164 x 10-4 m2/s 2.124 x 10-9 m = 3.28 x 105 m/s
1
E6 = -13.6 eV x 62 = -0.38 eV
d) r7 = 0.059 nm x 72 = 2.891 nm
7
v7 = = 1.164 x 10-4 m2/s 2.891 x 10-9 m = 2.81 x 105 m/s
1
E7 = -13.6 eV x 72 = -0.28 eV
7. Find the photon energy and wavelength of the light emitted for each of the following transitions:
a) n = 6 to n = 2
b) n = 7 to n = 2
c) n = 7 to n = 6
E
c
Ephoton = Ei - Ef
f= h
=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
 = f = 7.54 x 1014 Hz = 3.98 x 10-7 m = 398 nm
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
8. Calculate the deBroglie wavelength for an electron in the second and third energy levels of a Bohr
atom Check that a whole number of wavelengths fit around the circumference of the orbit.
h h
= =
p mv
h h
6.626 x 10-34 J·s
a)  = p = mv = (9.11 x 10-31 kg)( 9.86 x 105 m/s) = 7.38 x 10-10 m = 0.738 nm
C2 = 2r2 = 2(0.236 nm) = 1.48 nm
C 1.48 nm
=
= 2.005 ≈ 2
 0.738 nm
h h
6.626 x 10-34 J·s
b)  = p = mv = (9.11 x 10-31 kg)( 6.57 x 105 m/s) = 1.11 x 10-9 m = 1.11 nm
C2 = 2r3 = 2(0.531 nm) = 3.34 nm
C 3.34 nm
=
= 3.009 ≈ 3
 1.11 nm
9. 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?
h
x·p ≥
2
h
6.626 x 10-34 J·s
p ≥
=
= 1.41 x 10-28 kg·m/s
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 it momentum?
h
x·p ≥
2
h
6.626 x 10-34 J·s
p ≥
=
= 2.34 x 10-28 kg·m/s
2·x 2·450 x 10-9 m
10. 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 only 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 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 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 they jump to so there will only be 36 photon energies and 36 emission lines produced by
electrons jumping between those levels