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Practice problems for Final exam PHYS117B.02
1. Blackbody radiation is emitted by all objects that have finite temperature.
We will consider a number of items that you are familiar with and deduce
the blackbody radiation they emit.
(a) YOU! Your temperature is 37 o C, or 310 K. Find the peak wavelength and frequency of light emitted by you. What range of the
spectrum (radio, microwave, infrared, visible, ultraviolet, xray, gammaray) is this light found in?
The necessary equation is λm T = 2.9 × 10−3m × K. So, for a human
the peak wavelength is:
λm 310 = 2.9 × 10−3
2.9 × 10−3
λm =
310
λm = 9.35 × 10−6 m
(1)
(2)
(3)
This is microwave/infrared region.
(b) Your stove coil turns red hot. Take 600 nm as the peak wavelength
of this light. What is the temperature of the stove coil?
λm T = 2.9 × 10−3
600 × 10
−9
−3
T = 2.9 × 10
2.9 × 10−3
T =
600 × 10−9
T = 4833K
(4)
(5)
(6)
(7)
(c) Penzias and Wilson discovered that radio static is actually blackbody
radiation from the cold outreaches of space. They set the temperature
of space at 2.7 K (don’t fret over this, space is not completely empty
so it can have temperature). What is the peak frequency of the
blackbody radiation from cold space?
λm T = 2.9 × 10−3
(8)
−3
(9)
λm 2.7 = 2.9 × 10
2.9 × 10−3
λm =
2.7
λm = 1.07 × 10−3 m
λm f = c
c
3.00 × 108
f=
=
λm
1.07 × 10−3
f = 2.79 × 1011 Hz
1
(10)
(11)
(12)
(13)
(14)
2. A Helium atom consists of it’s nucleus (42 He) and 2 electrons. The energy
levels in this nucleus are not precisely the same as though in the Bohr
formula since the electrons are not only affected by the Z=2 nucleus, but
they are also affected by each other.
However, if one of the electrons from a He atom is removed, this positively
charged ion (a 42 He nucleus plus only ONE electron), is indeed quite accurately described by the Bohr formula. We will calculate a number of
things for this Z=2 ion. All we need to do is multiply the energy level
formula by Z 2 and then go on as for the hydrogen atom.
(a) If the electron is in the ground state, what energy photon is necessary
to remove it?
2
The basic equation for this problem is En = −13.6eV Zn2 . To remove
an electron from the ground state, you must make the transition from
n = 1 to n = ∞.
Ephoton = E∞ − E1
2
(15)
2
Z
Z
− (−13.6eV 2 )
2
∞
1
22
Ephoton = 0 − (−13.6eV 2 )
1
Ephoton = 54.4eV
Ephoton = −13.6eV
(16)
(17)
(18)
(b) If the electron is in the n=2 state, what energy photon is necessary
to remove it?
Ephoton = E∞ − E2
2
(19)
2
Z
Z
− (−13.6eV 2 )
∞2
2
22
Ephoton = 0 − (−13.6eV 2 )
2
Ephoton = 13.6eV
Ephoton = −13.6eV
(20)
(21)
(22)
(c) What energy photon is released if the electron drops from the n=3
to the n=2 state?
Ephoton = E3 − E2
Z2
Z2
Ephoton = −13.6eV 2 − (−13.6eV 2 )
3
2
22
22
Ephoton = −13.6eV 2 − (−13.6eV 2 )
3
2
2
(23)
(24)
(25)
Ephoton = −13.6eV
4
−1
9
5
9
= 7.55eV
Ephoton = 13.6eV
Ephoton
3
(26)
(27)
(28)
3. No two electrons can occupy identically the same quantum state. For this
reason, not all electrons on a given atom can be in the n=1 state, some of
them must be in higher states. As one considers bigger and bigger atoms,
the n=1 level will fill up, followed by n=2, followed by n=3, and so on...
(a) We will consider that atom that completely fills the n=2 level. Each
electron in this atom will have a set of 4 quantum numbers (n, l, ml , ms ).
Please fill in the table below specifying all four quantum numbers for
each of these electrons:
n l m l ms
1 0
0
+ 12
1 0
0
- 21
2 0
0
+ 12
2 0
0
- 21
2 1 -1 + 21
2 1 -1
- 21
2 1
0
+ 12
2 1
0
- 21
2 1 +1 + 12
2 1 +1 - 21
(b) What element is this in the periodic table?
This is 10 electrons and so this is Neon.
(c) What special property results from having the n=2 level completely
full?
Neon is a noble gas. It is not chemically active.
4
4. Uncertainty principle and de Broglie waves
(a) An atom in a metastable state has lifetime of 5.2 ms. What is the
uncertainty in the energy of this unstable state?
We use the uncertainty principle which relates the time for a cetratin
quantum process to occur and the energy involved in the process. In
this case we have an atom in a meta-stable state. We are given the
average lifetime of this state. It is fair to assume that we know this
time with uncertainty equal to this lifetime. Then we can find the
corresponding uncertainty in the energy of this state:
∆E∆t ≥ h̄
h̄
∆E ≥
∆t
h̄
∆E ≥
∆t
6.626 × 10−34 J.s
∆E ≥
2π5.2 × 10−3 s
−32
∆E ≥ 2.03 × 10 J = 1.27 × 10−13 eV
(29)
(30)
(31)
(32)
(33)
This is a rather long-lived state, so the uncertainty of the energy of
the state is negligible.
(b) A photon and an electron each have a wavelenth of λ = 250 nm. Calculate the energy of each. You want to study an organic molecule that
is about that long using either an electron or a photon microscope.
What wavelength should you use and which probe, the electron or
the photon, is likely to damage the molecule less?
Ephoton = hf
(34)
Ephoton = hc/λ
6.626 × 10−34 3 × 108
Ephoton =
eV
250 × 10−9 1.6−19
Ephoton = 4.97eV
p2
Ee =
2me
h
p=
λ
h2
Ee =
2me λ2
−34 2
(6.626 × 10 )
Ee =
J
2 × 9.1 × 10−31 .(250 × 10−9 )2
(35)
Ee = 3.8 × 10−24 J
Ee = 2.4 × 10−5 eV
5
(36)
(37)
(38)
(39)
(40)
(41)
(42)
(43)
To study the molecule you need a wavelength which is comparable ( or
smaller ) than the molecule size. From this point of view both photons
and electrons will do the job. However, the photon with λ = 250 nm has a
lot more !! energy than an electron of the same wavelength. The photon
is likely to damage the molecule, while the electron is not.
6
5. The ground state of the Li atom is the 2s level. The energies of the first
few excited states above the ground state are as follows:
level
2p
3s
3p
3d
4s
energy above ground state
1.85 eV
3.37 eV
3.83 eV
3.88 eV
4.34 eV
(a) Draw an energy level diagram and show the allowed transitions in
the emission spectrum of Li.
ANSWER: For a transition to be allowed, we need to follow the
selection rules: ∆l = 1 and ∆ml = 0, ±1. Since the atom is not in
external magnetic field, all levels that have the same angular momentum quantum number (l) have the same energy, i.e. at this point,
we don’t need to worry about ml . We can go from l = 0 ( s-states)
to l = 1 (p-states) (or vice versa) and from l = 1 (p-states) to l = 2
(d-states) or vice versa. This is emission spectrum, so we can have all
combinations of the given levels that obey the selection rules. That
gives:
7
2p - 2s
3s - 2p
3p - 2s
3p - 3s
3d - 2p
3d - 3p
4s - 2p
(b) Which of these lines are present in the absorption spectrum? ANSWER: Now, for absorption, we have to start from the atom being
in the ground state. So the initial state is 2s. Then we can only have:
2s - 2p
2s - 3p
(c) Calculate the wavelengths of the absorption lines and indicate whether
they are in the infrared,visible or ultraviolet range.
Ephoton = E2p − E2s
Ephoton = 1.85eV
(44)
(45)
λ = hc/Ephoton = 670nm
(46)
This is visible light.
Ephoton = E3p − E2s
Ephoton = 3.83eV
(47)
(48)
λ = hc/Ephoton = 324nm
(49)
This is in the ultra-violet range.
(d) If you place the atom in an external magnetic field, the spectral lines
will get split into several lines due to the interaction between the
orbital angular momentum of the electrons with the magnetic field
(ignore the spin). Pick one of the transitions that you identified in (a)
and expalain how the energy levels will split and how many spectral
lines will appear as a result of this splitting.
Let’s pick 3d - 3p. The upper level will split into 5 levels with
ml = −2, −1, 0, 1, 2, the lower level will split into 3 levels with
ml = −1, 0, 1. All of these levels are equally spaced by ∆E = ml µb B,
eh̄
is Bohr’s magneton. In
where B is the magnetic field and µb = 2m
e
transitions, we now need to observe the selection rule ∆m = 0, ±1,
so not all transitions are allowed. From ml = 2 we can only go to
ml = 1, from ml = 1 we can go to ml = 1or0 , etc...We note that
the transitions ml = 2 to ml = 1, ml = 1 to ml = 0, ml = 0 to
ml = −1 all have the same wavelength since the energy levels are
equally spaced. Thus we find that all transition for which ∆m = −1
have the same wavelength. The same is true for transitions with
∆m = 0 or ∆m = 1. As a result, there are only 3 different wavelengths arising from the 3d - 3p line.
8
6. Some particular material has a work function of 5 eV. For this material:
(a) Calculate the lowest frequency photon that can remove an electron
from this material.
E = hf
E
f=
h
6.626 × 10−34
eV s
1.6 × 10−19
h = 4.141 × 10−15 eV s
5eV
f=
4.141 × 10−15 eV s
f = 1.21 × 1015 Hz
h = 6.626 × 10−34 Js =
(50)
(51)
(52)
(53)
(54)
(55)
(b) Calculate the wavelength of a photon that removes electrons from the
material such that the electrons leave with 8 eV as their maximum
kinetic energy.
OK, since the work function is 5 eV, and the electrons leave with 8
eV, the photon must supply a total energy:
Ephoton = 5eV + 8eV = 13eV
hc
Ephoton = 13eV = hf =
λ
hc
= 13eV
λ
hc
λ=
13eV
4.141 × 10−15 ∗ 3.0 × 108
λ=
13eV
λ = 9.55 × 10−8 m = 95.5nm
9
(56)
(57)
(58)
(59)
(60)
(61)
7. Close study of some heavy nuclei showed that they were not stable and
emitted “rays”. This was different from X-rays (that were purposely made
by man) since these special elements voluntarily made radiation instead of
by conversion of the kinetic energy of the electrons into photons. Different
types of these rays were labeled with the Greek letters α, β, andγ. We now
know better what these rays are:
(a) Identify what each of the following are made of:
i. alpha rays. Helium nuclei.
ii. beta rays.
electrons
iii. gamma rays.
photons
(b) In free space, a neutron has a lifetime of only 10 minutes! Write
down the equation that shows the decay of a neutron. Your equation
should look like “n → something + something + something”.
n → p + e− + ν¯e
(c) In a 12
6 C nucleus, the neutrons cannot decay, because all available
states for the produced proton are already filled. But, in a 14
6 C nucleus, the “last” neutron can decay. Write an equation that describes
this decay in the form “14
6 C → something + something + something”.
14
14
−
+ ν¯e
6 C → 7 N + e
(d) Would the decay listed in the previous step be considered alpha, beta,
or gamma decay?
It is beta minus decay, because the detected particle is the electron.
(e) A proton in free space does not decay. However, in the nucleus 11
6 C
it can indeed decay. Write an equation for this decay in the form
“11
6 C → something + something + something”.
11
11
+
+ νe
6 C → 5 B + e
(f) Living material maintains the ratio Rliving =
14
6 C
has a 5730 year half-life, while
12
6 C
14
6 C
12 C
6
∼ 1.3 × 10−12 .
is stable.
14
6 C
12 C
6
i. Calculate the
ratio for a sample of material that is 2,000
years old.
Since 12
6 C is stable, its amount in the sample will not change
after the death of the living organism, but the 14
6 C will get less
and less over time, hence the ratio of the two will decrease over
time following the radioactive decay law.
14
6 C
12 C
6
λ=
10
= R0 e−λt
(62)
ln(2)
0.693
=
t 12
t 12
(63)
λ=
0.693
1
= 1.21 × 10−4
5730years
years
14
C
6
−12 −1.21×10−4 t
e
12 C = 1.3 × 10
6
(64)
(65)
In this case, we can calculate the remaining carbon using 2000
for t:
14
6 C
12 C
6
14
6 C
12 C
6
t
(66)
2000
(67)
= 1.3 × 10−12 e−1.21×10
= 1.3 × 10−12 e−1.21×10
−4
−4
14
6 C
−12 −0.242
e
12 C = 1.3 × 10
6
14
6 C
−12
0.785
12 C = 1.3 × 10
6
14
6 C
−12
12 C = 1.02 × 10
6
ii. Calculate the
years old.
14
6 C
12 C
6
= 1.3 × 10−12 e−1.21×10
14
6 C
12 C
6
(71)
−4
20000
(72)
−4
(73)
(74)
(75)
ratio for a sample of material that is 65 Million
14
6 C
12 C
6
14
6 C
12 C
6
(70)
t
= 1.3 × 10−12 e−1.21×10
14
6 C
−12 −2.42
e
12 C = 1.3 × 10
6
14
6 C
−12
0.08892
12 C = 1.3 × 10
6
14
6 C
−12
12 C = 0.1156 × 10
6
iii. Calculate the
years old.
(69)
ratio for a sample of material that is 20,000
14
6 C
12 C
6
14
6 C
12 C
6
(68)
t
(76)
65M y
(77)
= 1.3 × 10−12 e−1.21×10
= 1.3 × 10−12 e−1.21×10
11
−4
−4
14
6 C
12 C
6
14
6 C
12 C
6
= 1.3 × 10−12 e−7865
(78)
= ZERO(calculatorf ails)
(79)
iv. Explain why Carbon dating is not used for dinosaur bones.
There is too little of of the original 14
6 C left to be measured.
(g)
229
90 T h
(Thorium) decays by an alpha decay. Write an equation for
this decay in the form “229
90 T h → something + something”. The
following information might be helpful:
(86 Rn, 87 F r, 88 Ra, 89 Ac, 90 T h, 91 P a, 92 U, 93 N p).
229
4
225
90 T h → 2 He + 88 Ra
12
8. Calculate all the following:
(a) The radius of a Gold (197
79 Au) nucleus.
1
(80)
r = r0 A 3
1
3
(81)
(82)
r = 1.2f m197
r = 6.98f m
(b) The radius of an Alpha (42 He) particle.
Note: This formula is poor for the He nucleus, but still helps to make
the point to where I want to get to next with this problem.
1
(83)
r = r0 A 3
1
3
r = 1.2f m4
r = 1.9f m
(84)
(85)
(c) Calculate the energy of the alpha particle necessary to bring it so
close to a Au nucleus that the edges touch (i.e. distance between
nuclei = RAu + RHe ).
Q1 Q2
r
Q1 = 2 ∗ 1.6 × 10−19 = 3.2 × 10−19 C
Q1 = 79 ∗ 1.6 × 10−19 = 126.4 × 10−19 C
3.2 ∗ 126.4 × 10−38
U = 9 × 109
J
(1.9 + 7.0) × 10−15
U =k
U = 4.1 × 10−12 J
4.1 × 10−12
U=
eV
1.6 × 10−19
U = 25.6 × 106 eV = 25.6M eV
(86)
(87)
(88)
(89)
(90)
(91)
(92)
Note: to get so close to the Au nucleus, the alpha particle (2 charges)
would need to be accelerated through a potential difference of 12.8
million volts!
(d) Explain what happens to the alpha particle if it comes closer to the
Au nucleus than the result calculated in the previous step.
The alpha particle will feel the nuclear force and be absorbed by the
nucleus.
(e) What new force was introduced into physics to explain the fate of
the high energy alpha particle.
The nuclear force or the strong force.
13
9. Inside a nucleus, the nuclear force attracts (strongly) but the electric force
repels (weakly). The nuclear binding energy curve is shown in the figure
below:
(a) Explain, using the characteristics of the nuclear and electric forces
why this curve rises then falls.
The curve rises because the nuclear attraction makes larger nuclei
more tightly bound. However, eventually the unlimited range electric
force wins over the finite range nuclear force and nuclei become less
bound. Above iron nuclei, the binding energy curve falls.
(b) Explain, using the characteristics of the nuclear and electric forces,
why the rise is steeper than the fall.
This rise is driven by the strong nuclear force. The fall is driven by
the weaker electric force. This is why the rise is steep and the fall is
not.
(c) To release energy from a Uranium nucleus, would you use fission or
fusion? Explain using the nuclear binding energy curve.
Uranium nuclei are on the right of the peak in the binding energy-pernucleon curve. Thus, if we use fusion, we’ll make a bigger nucleus
which will be less tightly bound - no way to release energy. If we
split the Uranium nucleus in two parts, we’ll gain energy, because
the daughter nuclei will be more tightly bound.
(d) To release energy from two deuterons (21 H), would you use fission or
fusion? Explain using the nuclear binding energy curve.
Fusion will bring us to a more tightly bound nucleus, thus we’ll gain
energy.
14
(e) One atomic mass unit is u = 1.6605402 × 10−27 kg. The rest mass
of a deuteron 21 H is 2.0141018 u and the rest mass of a helium 42 He
nucleus is 4.026032 u. Calculate the energy released in the reaction
2
2
4
1 H + 1 H → 2 He.
Two deuterons account for a mass more than the Helium
Mdeut = 2 ∗ 2.0141018u = 4.0282036u
(93)
MHe = 4.026032u
Mdeut − MHe = 0.0021716u
(94)
(95)
1u = 931.5M eV
931.5M eV
= 2.022M eV
E = 0.0021716u
1u
(96)
(97)
(f) The result from the previous step is the energy released by a single
nuclear reaction. What is the energy released is 2 moles of 21 H were
used (i.e. producing 1 mole of 42 He).
Etot = E1 ∗ NA = 2.022M eV ∗ 6 × 1023
Etot = 1.21 × 1024 M eV = 1.21 × 1030 eV
(98)
(99)
Etot = 1.21 × 1030 1.6 × 10−19 J
Etot = 1.94 × 1011 J
(100)
(101)
WOW!
(g) What is the environmental impact of He?
Nothing.
15
10. The “rule of thumb” for an NMR device is that protons will achieve resonance at a frequency of 42.58 MHz when they are immersed in a magnetic field of 1.0 Tesla. Assume that a patient is placed in a magnetic
field, B, whose strength varies with position, x, according to the formula
B(x) = 1.0 T esla + 0.1 T esla
m x.
(a) Especially strong resonant signals are received coming back from the
patient for frequencies of 43.00 MHz and 45.00 MHz. Find the locations in x that produce these signals.
The external magnetic field makes the energy levels in the nucleus
split according to their spin. Thus when nuclei flip their spin from
one direction to another, they will jump between energy levels and
emit (or absorb) photons. If we want to obtain position information
on where exactly the emission happened in the patient’s body, we
want to make the energy level splitting a little bit different in each
position in the body, i.e. we want each part of the body to send us a
different frequency signal or “play a different note”. We can achieve
that by applying a magnetic field that varies with distance.
So if the main frequency is f0 = 42.58 MHz, this is achieved at
B0 = 1T . The energy level splitting is proportional to the strength
of the magnetic field.I can write:
E0 = hf0 = kB0
(102)
,where k is some constant which depends on the magnetic quantum
number and the nuclear magneton.
E1 = hf1 = kB1 = k(B0 + 0.1x1 )
E2 = hf2 = kB2 = k(B0 + 0.1x2 )
f1
0.1
=1+
x1
f0
B0
0.1T m
43.00M Hz
=1+
x1
42.58M Hz
1T
1.00986 = 1 + 0.1x1
x1 = 0.0986m = 9.86cm
0.1
f2
=1+
x2
f0
B0
45.00M Hz
0.1T m
=1+
x2
42.58M Hz
1T
1.05683 = 1 + 0.1x2
x2 = 0.568m = 56.8cm
(112)
16
(103)
(104)
(105)
(106)
(107)
(108)
(109)
(110)
(111)
(b) The doctor can think of this technique as one that “maps the location
of water” inside the patient. Explain why.
The NMR signal is strongest from the proton. This is most often
found in the body as one of the H molecules in H2 O.
17
11. PET scanning is an additional method of medical imaging that involves
the annihilation of a positron inside a patient producing two photons. The
mass of a positron is the same as the mass of an electron, me = 9.109 × 10−31 kg.
(a) What is the mass energy of the positron?
E = 9.109 × 10
−31
E = mc2
2
kg 3.0 × 108
8.19 × 10
eV
1.6 × 10−19
E = 511000eV = 511keV
(113)
(114)
−14
E=
(115)
(116)
(b) How much energy is released during the annihilation process?
The mass energy of the electron AND the positron are released:
Etot = Ep = 2 ∗ 511keV = 1022keV
(117)
(c) This energy is divided equally between the two photons. What is the
frequency of each of these photons?
Ephoton = 511keV = hf
1e
h = 6.626068 × 10−34 Js
1.602 × 10−19 C
h = 4.135667 × 10−15 eV s
E
511000
f=
=
= 1.2 × 1020 Hz
h
4.135667 × 10−15
18
(118)
(119)
(120)
(121)
12. Shown in the picture below is an X-ray tube.
(a) Where do the electrons come from? The hot filament is able to release
electrons (the most energetic ones).
(b) Why are the electrons pushed to the right?
The filament is negative and the W/Cu target is positive. So, the
electric force pushes them.
(c) What physical process causes the electron energy to turn into EM
radiation?
Bremsstrahlung or “braking radiation”. The electrons slow down
when passing through matter ... then the lost eregy turns into radiation energy, which happens to be in the X-ray range.
(d) Suppose the accelerating voltage of the X-ray tube were 70,000 Volts.
i. What is the energy of each electron as it strikes the tungsten/copper
plate?
U = qV
(122)
q = 1e
U = 1e ∗ 70000V = 70000eV
(123)
(124)
U = 70keV
(125)
ii. Assuming that all this energy goes into a single photon, what is
the frequency of that photon?
19
Ephoton = hf
(126)
6.626068 × 10−34
h=
eV s
1.602 × 10−19
(127)
h = 4.135667 × 10−15 eV s
E
70000
f=
=
= 1.69 × 1019 Hz
h
4.135667 × 10−15
(128)
(129)
iii. What is the wavelength of that photon?
λf = c
3.0 × 108
c
λ= =
f
1.69 × 1019
λ = 1.77 × 10−11 m
20
(130)
(131)
(132)