In the semiconductor laser, electrons are optically pumped from the valence band to the conduction band with a light source that produces photons of frequency v P and energy E P = hv P,
Trang 1PHYSICS TOPICAL:
Atomic and Nuclear Structure
Test 1
Time: 23 Minutes*
Number of Questions: 18
* The timing restrictions for the science topical tests are optional If you are using this test for the sole purpose of content reinforcement, you may want to disregard the time limit
Trang 2DIRECTIONS: Most of the questions in the following test
are organized into groups, with a descriptive passage preceding each group of questions Study the passage, then select the single best answer to each question in the group Some of the questions are not based on a descriptive passage; you must also select the best answer
to these questions If you are unsure of the best answer, eliminate the choices that you know are incorrect, then select an answer from the choices that remain Indicate your selection by blackening the corresponding circle on your answer sheet A periodic table is provided below for your use with the questions
PERIODIC TABLE OF THE ELEMENTS
1
H
1.0
2
He
4.0 3
Li
6.9
4
Be
9.0
5
B
10.8
6
C
12.0
7
N
14.0
8
O
16.0
9
F
19.0
10
Ne
20.2 11
Na
23.0
12
Mg
24.3
13
Al
27.0
14
Si
28.1
15
P
31.0
16
S
32.1
17
Cl
35.5
18
Ar
39.9 19
K
39.1
20
Ca
40.1
21
Sc
45.0
22
Ti
47.9
23
V
50.9
24
Cr
52.0
25
Mn
54.9
26
Fe
55.8
27
Co
58.9
28
Ni
58.7
29
Cu
63.5
30
Zn
65.4
31
Ga
69.7
32
Ge
72.6
33
As
74.9
34
Se
79.0
35
Br
79.9
36
Kr
83.8 37
Rb
85.5
38
Sr
87.6
39
Y
88.9
40
Zr
91.2
41
Nb
92.9
42
Mo
95.9
43
Tc
(98)
44
Ru
101.1
45
Rh
102.9
46
Pd
106.4
47
Ag
107.9
48
Cd
112.4
49
In
114.8
50
Sn
118.7
51
Sb
121.8
52
Te
127.6
53
I
126.9
54
Xe
131.3 55
Cs
132.9
56
Ba
137.3
57
La *
138.9
72
Hf
178.5
73
Ta
180.9
74
W
183.9
75
Re
186.2
76
Os
190.2
77
Ir
192.2
78
Pt
195.1
79
Au
197.0
80
Hg
200.6
81
Tl
204.4
82
Pb
207.2
83
Bi
209.0
84
Po
(209)
85
At
(210)
86
Rn
(222) 87
Fr
(223)
88
Ra
226.0
89
Ac †
227.0
104
Rf
(261)
105
Ha
(262)
106
Unh
(263)
107
Uns
(262)
108
Uno
(265)
109
Une
(267)
*
58
Ce
140.1
59
Pr
140.9
60
Nd
144.2
61
Pm
(145)
62
Sm
150.4
63
Eu
152.0
64
Gd
157.3
65
Tb
158.9
66
Dy
162.5
67
Ho
164.9
68
Er
167.3
69
Tm
168.9
70
Yb
173.0
71
Lu
175.0
†
90
Th
232.0
91
Pa
(231)
92
U
238.0
93
Np
(237)
94
Pu
(244)
95
Am
(243)
96
Cm
(247)
97
Bk
(247)
98
Cf
(251)
99
Es
(252)
100
Fm
(257)
101
Md
(258)
102
No
(259)
103
Lr
(260)
GO ON TO THE NEXT PAGE.
Trang 3
Passage I (Questions 1–6)
Semiconductor lasers typically produce light with a
wavelength on the order of 1 µm The active region of the
laser consists of a thin, narrow semiconductor crystal about
200 µm long with partially reflecting facets at the ends
through which light can enter or exit
In a semiconductor crystal, electron energies are
confined to bands (collections of energy levels very close
together in energy) rather than to the discrete energy levels
found in a single atom All semiconductors have a valence
band, which is almost completely filled with electrons, and
a higher energy conduction band The smallest energy
difference between these two bands is called the band gap
E g Figure 1 shows the energy of the bands, E, as a
function of the electron wave momentum for a typical
semiconductor (k is the magnitude of the wave vector of
the electron and plays a role analogous to that of a quantum
number: the larger its value, the greater the momentum of
the electron.) In the semiconductor laser, electrons are
optically pumped from the valence band to the conduction
band with a light source that produces photons of frequency
v P and energy E P = hv P, where h is Planck’s constant
k
E(k)
E g
Conduction Band
Valence Band
Figure 1
A second light source, having energy E s near the band
gap, can now be amplified by the process of stimulated
emission In this process, a photon interacts with a
conduction band electron causing it to drop to the valence
band This induces the emission of another photon of the
same frequency, which then stimulates another conduction
band electron to drop to the valence band The number of
photons increases, and the light is amplified as it travels
back and forth through the active region Figure 2 shows a
graph of the natural log of the output intensity I out divided
by the input intensity I in as a function of E s for a typical
semiconductor laser (Note: hc = 1.24 × 10–6 eV•m, where
c is the speed of light in a vacuum.)
0
s
g
/Iin
Figure 2
1 Semiconductor lasers are often used as light sources for
optical fiber communications As a photon travels from the laser (index of refraction 3.5) into a glass fiber (index of refraction 1.4), which of the following changes?
I Its energy
II Its frequency III Its wavelength
A I only
B III only
C I and II only
D I, II, and III
2 Which of the following is true of the energy E p of the photons produced by the optical pump?
A E p must be smaller than the band gap E g
B E p must equal the band gap E g
C E p must be greater than or equal to the band gap
E g
D E p can be chosen independently of the band gap
E g
3 What value of E s will yield the most efficient laser?
A the value just before the vertical drop in Figure 2
B the value just after the vertical drop in Figure 2
C E g
D infinity
4 A photon emitted by a semiconductor laser is incident
on a hydrogen atom Assuming no ionization takes place, the photon will be absorbed by the atom:
A only if its energy equals the difference between
two electron energy levels
B only if its energy is an integer multiple of
Planck’s constant h
C only if its energy is greater than 13.6 eV.
D independent of the value of its frequency.
GO ON TO THE NEXT PAGE.
Trang 45 A particular semiconductor has a band gap of 2.48 eV.
What is the wavelength of the photon that is emitted
when an electron drops from the lowest energy state of
the conduction band to the highest energy state of the
valence band?
A 4.0 µm
B 2.0 µm
C 1.0 µm
D 0.5 µm
6 Examination of Figure 2 reveals that above a certain
energy, ln (Iout/Iin) becomes less than 0 This implies
that:
A E p = E g
B E g = 0
C the input light is being absorbed instead of
amplified
D the process of stimulated emission is amplifying
the input light
GO ON TO THE NEXT PAGE.
Trang 5
Passage II (Questions 7–13)
Figure 1 shows a graph of the logarithm of the
relative abundance of the elements in the universe as a
function of mass number A The dominant process for the
synthesis of elements heavier than iron (Fe) is neutron
capture In neutron capture, energetic neutrons bombard a
target nucleus which absorbs one or more of them to form
an unstable isotope The unstable nucleus then undergoes
beta decay until it becomes stable A nucleus produced by
one or more beta decays is called a beta decay daughter
There are two situations in which neutron capture
occurs The first case is called the s-process and results
when the neutron density of the environment is moderate
The capture process is slow, and the time that it takes the
target nucleus to absorb a neutron is large compared to the
time that it takes for the resulting unstable nucleus to beta
decay Certain nuclei with a “magic” number of neutrons
have a small probability for capture Therefore, beta decay
daughters with magic neutron numbers N = 50, 82, and
126 account for the abundance peaks in Figure 1 at A = 90,
138, and 208
The second situation in which neutron capture occurs
is called the r-process and results when the neutron density
of the environment is very high In this case the rate of
capture is very rapid compared to the beta decay rate of the
resulting unstable nucleus Therefore, a nucleus captures
many neutrons before undergoing beta decay Since an
unstable nucleus with a magic number of neutrons has a
low probability of absorbing more neutrons, its beta decay
daughter will have a relatively high abundance This
accounts for the second set of peaks in Figure 1 at A = 80,
130, and 194
10
8
6
4
2
0
-2
50 100 150 200 Atomic Weight
Iron Group
N= 50 N=82 N=126 Li-Be-B
H He
D
Figure 1
Figure 1 adapted from Reviews of Modern Physics Journal #29, 1957.
Reprinted courtesy of the American Physical Society.
7 11849In undergoes a beta decay in the following reaction: 11849In→X + e–.What is the element X?
A 11949In
B 11850Sn
C 11848Cd
D 11H
8 Although the Sun is primarily composed of hydrogen,
why is rapid neutron capture more common than rapid proton capture?
A Free neutrons rapidly decay to stable protons.
B Neutrons are in greater supply than protons.
C Free protons have less kinetic energy than free
neutrons because they are heavier
D Neutrons do not experience electrostatic repulsion
from the nucleus
9 Which of the following nuclear reactions will
maximize the production of heavy elements by neutron capture?
A 6 13 2 4 8 16
C+ He→ O+n
B 1 1 1 2 2 3
H+ H→ He
C 5 8 4 8
B→ Be + e+
D n→ +p e−
1 0 A 11148Cd atom is exposed to a high density of neutrons and absorbs 7 of them before it beta decays After the neutron flux discontinues, what is the final stable isotope produced?
A 11848Cd
B 12250Sn
C 11850Sn
D 11949In
GO ON TO THE NEXT PAGE.
Trang 61 1 What is the ratio of the abundance of hydrogen in the
universe to that of iron?
A 1:104
B 1:103
C 105:1
D 106:1
1 2 An element undergoes beta decay, and 25% of the
original sample remains after 16 hours What is the
half-life of the element?
A 1 hour
B 2 hours
C 4 hours
D 8 hours
1 3 A stable nucleus resulting from an s-process neutron
capture is likely to:
A have a magic number of neutrons.
B result from a single beta decay of a nucleus having
a magic number of neutrons
C beta decay into a nucleus having a magic number
of neutrons
D undergo an r-process neutron capture.
GO ON TO THE NEXT PAGE.
Trang 7
Questions 14 through 18 are NOT
based on a descriptive passage
1 4 The element 13245∆ is formed as a result of 2 alpha
decays, 2 positron decays and a gamma decay Which
of the following is the parent element?
A 12439Λ
B 14047Γ
C 14049Φ
D 14051Ω
1 5 In demonstrating the photoelectric effect with zinc,
ultraviolet light is used rather than visible light
because ultraviolet light:
A moves faster.
B has a greater energy per photon.
C is more intense.
D is not visible.
1 6 The four lowest energy levels of an electron in an
atom are the ground state and the first three excited
states What is the maximum number of different lines
in the emission spectrum of the atom that can be
accounted for by transitions in these four levels?
A 3
B 4
C 6
D 12
1 7 Four hydrogen atoms fuse to form a helium atom in a
series of reactions How much energy, in MeV, is
released in this process? (Note: The atomic mass of
hydrogen is 1.0080 amu and the atomic mass of
helium is 4.0026 amu c2 = 932 MeV/amu.)
A 0 MeV
B 6.8 MeV
C 18.9 MeV
D 27.4 MeV
1 8 The following is a table of the abundance of a
radioactive element as a function of time What is the half-life of the element?
radioactive particles (10 23 )
A 2.5 years
B 3.5 years
C 5 years
D Cannot be determined from the information given.
END OF TEST
Trang 8ANSWER KEY:
Trang 9
EXPLANATIONS
Passage I (Questions 1—6)
The photon, being a quantum of light, exhibits also the wave behavior of light (and of waves in general) as it crosses the boundary from one medium to another The frequency remains the same, and so does the energy since it is related to the frequency by the equation E = hf The wavelength and the speed, on the other hand, do change The speed of light in a medium with an index of refraction n ( 1) is given by v = c/n, where c is the speed in vacuum This accounts for the refraction of light
as described by Snell’s law Since the speed changes while the frequency remains constant, the wavelength will have to change
to maintain the equality v = λf Only statement III is therefore correct, making B the correct choice
In the passage we are told that electrons are pumped from the valence band to the higher energy conduction band The band gap Eg is the smallest difference in energy between these two bands When an electron is optically pumped, it absorbs a photon of a particular energy and makes a transition to a higher energy state By conservation of energy, the difference between the final and initial energy states of the electron equals the energy of the photon absorbed Within the band gap there are no states that the electron can occupy, so the electron must absorb enough energy to make a transition across the gap The smallest energy that can bring about this transition is equal to the band gap In this case an electron in the highest energy level of the valence band is excited to the lowest energy level of the conduction band In general, an electron may start at a lower level in the valence band and/or end up at a higher energy level in the conduction band The photon energy then will need to be greater The photon absorbed by an electron must therefore have an energy at least as big as the band gap in general
Eg(band gap) valence band
conduction band
transition
Even though efficiency is not formally defined for us in this context, it should not be difficult to reason that the higher the output intensity relative to the input intensity, the more efficient the system is in extracting energy Either from this alone
or from examining the answer choices, our attention should be directed to an examination of the graph in Figure 2 The x-axis corresponds to the energy of the second light source, while the y-axis plots the natural logarithm of Iout/Iin The properties of logarithms are worth reviewing In general, the larger a number, the larger its natural logarithm, but the two do not increase at the same rate The natural logarithm of a negative number is undefined; the natural logarithm of a number between 0 and 1 is negative; while the natural logarithm of 1 is zero From the graph, the y-value is zero when Es < Eg So under such conditions, the quotient (Iout/Iin) is equal to one, since its natural logarithm is zero Or, in other words, Iout = Iin The y-value then increases until the vertical drop Immediately before the drop, the y-value is at its maximum, and therefore so is Iout/Iin: the laser is at its most efficient Immediately after the drop, the y-value takes on a negative value (and stays negative) A negative value for the natural logarithm means that the quotient (Iout/Iin) is less than one: i.e., the output intensity is less than the input intensity
In short, the laser is operating at its most efficient level (producing the most output intensity per unit of input intensity) when Es takes on the value right before the discontinuity occurs
When a hydrogen atom absorbs a photon, its electron makes a transition to a higher energy state This can only occur
if the difference in energy between the initial and final states equals the energy of the photon Choice B states that the photon has to have an energy that is an integer multiple of h We know that the energy of a photon is E = hf, where f is the frequency Choice B then suggests that the photon’s frequency has to be an integer in order for it to be absorbed This is a gross distortion
of the concept of quantization The actual quantization requirement is that the energy of light comes in packets and has a minimum value of hf, i.e it can be hf, 2hf, 3hf, etc There is no restriction on the value of f itself Besides, this requirement tells us nothing about the absorption of the photon Choice C is the condition for ionization: A photon above this value will eject the electron from the atom, and the residual energy will be the kinetic energy of the electron We are told, however, that ionization does not occur and besides, a photon can be absorbed without causing ionization So choice C is incorrect Choice D
is also clearly wrong in that the energy of a photon is proportional to its frequency
Trang 105 D
The energy of the photon emitted is equal to the energy difference between the final and initial states of the electron The final state of the electron in this case is the highest energy state of the valence band, while the initial state is the lowest energy state of the conduction band The energy difference is the band gap (See also explanation to #2 above.) The photon energy must therefore be 2.48 eV To determine its wavelength, we use the relationship:
E = hf = hc/λ
λ = hc
E =
1.24 × 10–6 eV•m 2.48 eV = 0.5 × 10–6 m = 0.5 µm
As discussed in the explanation to #3, ln (Iout/Iin) < 0 means that (Iout/Iin) < 1 This is a property of the natural logarithmic function The output intensity is therefore less than the input intensity Some of the input light has been absorbed
Choice A states that the energy Ep of the photon that “pumps” electrons from the valence to the conduction band is equal to the band gap Eg This cannot be deduced: pumping is needed to populate the conduction band with electrons so they can relax As long as this is accomplished, the actual energy of photons used in pumping plays no role in the subsequent processes
of stimulated emission and/or absorption Choice B states that the band gap is zero This need not be the case either Choice D describes the region of the plot where ln (Iout/Iin) is greater than zero
Passage II (Questions 7—13)
This question does not require any information from the passage and is a straightforward one dealing with beta-decay In this process, a neutron decays into a proton and an electron that is ejected from the nucleus As a result, the mass number A (number of neutrons + protons) does not change The atomic number, Z, which is the number of protons in the nucleus, increases by 1 The general scheme for the beta-decay of an isotope Y is therefore:
Z
AY→Z+A1X+–10β
where the beta-particle is the ejected electron Given the parent nucleus in the question, the daughter nucleus has to be one with atomic number 49 + 1 = 50 and mass number 118
Neutrons and protons both reside in the nucleus and have roughly the same mass The most important difference is in the charge: The proton is positively-charged while the neutron carries no charge How does this come into play? In order for a particle to be captured by a nucleus, it must penetrate the nucleus and get close enough to the nucleons so that the short range strong force can exert its effect and bind the particle to the nucleus Since the protons in the original nucleus impart to it a positive charge, it will repel other protons that approach it A neutron, however, experiences no electrostatic repulsion from the nucleus, making the initial approach much easier
We are told in the first paragraph of the passage that in neutron capture, the target nucleus absorbs one or more neutrons with which it is being bombarded, and then undergoes beta-decay To initiate the process, then, a source of free neutrons is needed Among all the reactions given in the choices, only choice A produces neutrons
Cadmium-111, upon absorption of 7 neutrons, will become cadmium-118 However, this is not a stable nucleus and,
as the passage tells us, subsequently undergoes beta-decay We are not told that how many times beta-decay occurs; but the mass number should remain at 118 Only choices A and C are possible Choice A corresponds to no beta-decay, and therefore contradicts the information given Choice C is the daughter nucleus after two beta-decay processes, and is the correct answer Note that if one did not pay attention to the mass number, and assumed that beta-decay occurs only once, one might have leapt
to the wrong choice of D