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TIME
OF
COMPLETION_______________
NAME______SOLUTION_______________________
DEPARTMENT OF NATURAL SCIENCES
PHYS 3650, Exam 2
Version 1
Total Weight: 100 points
Section 1
October 31, 2005
1.
Check your examination for completeness prior to starting. There are a total of ten (10)
problems on seven (7) pages.
2.
Authorized references include your calculator with calculator handbook, and the Reference
Data Pamphlet (provided by your instructor).
3.
You will have 75 minutes to complete the examination.
4.
The total weight of the examination is 100 points.
5.
There are six (6) multiple choice and four (4) calculation problems. Work all problems.
Show all work; partial credit will be given for correct work shown.
6.
If you have any questions during the examination, see your instructor who will
be located in the classroom.
7.
Start:
Stop:
10:30 a.m.
11:45 a.m
PROBLEM
POINTS
1-6
30
7
20
8
20
9
15
10
15
TOTAL
100
PERCENTAGE
CREDIT
CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN
MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR
PARTIAL CREDIT.
1. The p state of an electronic configuration corresponds to
a. n = 2.
b. l = 2.
(5)
c.
l = 1.
d. n = 0.
2. Bohr’s quantum condition on electron orbits requires
a. That the angular momentum of the electron about the hydrogen nucleus equal
nh/(2).
b.
That no more than one electron occupy a given stationary state.
(5)
c. That the electrons spiral into the nucleus while radiating electromagnetic waves .
d. None of the above .
3. The total number of quantum states of hydrogen with quantum number n = 4 is
a. 4.
b. 16.
(5)
c. 32.
d. 36.
4. An electron in the L shell means that
a. l = 1.
b. n = 1.
(5)
c. n = 2.
d. m = 2.
5.
The restriction that no more than one electron can occupy a given quantum state in an
atom was first stated by which of the following scientists?
a. Bohr.
b. De Broglie.
(5)
c. Heisenberg.
d. Pauli.
6. If the principal quantum number for hydrogen is 5, which one of the following is not a
permitted orbital magnetic quantum number for that atom?
a. 6.
b. -2.
(5)
c. 0.
d. 3.
7. The optical spectra of atoms with two electrons in the same outer shell are similar, but they are
quite different from the spectra of atoms with just one outer electron because of the interaction of the
two electrons. Separate the following elements into two groups such that those in each group have
similar spectra: lithium (Z = 3), beryllium (Z = 4), sodium (Z = 11), magnesium (Z = 12), potassium
(Z = 19), calcium (Z = 20), chromium (Z = 24), nickel (Z = 28). Please list the electronic
configurations of all of these elements.
Li: 1s22s1
Be: 1s22s2
Na: 1s22s22p63s1
Mg: 1s22s22p63s2
K: 1s22s22p63s23p64s1
Ca: 1s22s22p63s23p64s2
Cr: 1s22s22p63s23p64s23d4
Ni: 1s22s22p63s23p64s23d8
1 electron in the outer shell: Li, Na, K
2 electrons in the outer shell: Be, Mg, Ca, Cr, Ni
8. Calculate the wavelength of the K line of rhodium (Z = 45).
K: transition from L shell to K shell
EK = - (Z-1)2 E0/12 = -26,330 eV
EL = - (Z-1)2 E0/22 = -6,582 eV
E = EL – EK = 19,748 eV
 = hc/E = 0.0629 nm
Calculate the wavelength of the L line of rhodium.
L: transition from M shell to L shell
EM = - (Z-9)2 E0/32 = -1,958 eV
EL = - (Z-1)2 E0/22 = -6,582 eV
E = EM – EL = 4,624 eV
 = hc/E = 0.269 nm
9. Spectral lines of the following wavelengths are emitted by singly ionized helium: 164 nm,
230.6 nm, and 541 nm. Identify the transitions that result in these spectral lines.
He: Z = 2
En = - Z2 E0/n2
E1 = -54.4 eV
E2 = -13.6 eV
E3 = -6.04 eV
E4 = -3.40 eV
E5 = -2.18 eV
E6 = -1.51 eV
E7 = -1.11 eV
E8 = -0.850 eV
E9 = -0.671 eV
E10 = -0.544 eV
 = hc/E
E= hc/
E1= hc/ (1243 eVnm)/(164 nm) = 7.58 eV
E2= hc/ (1243 eVnm)/(230.6 nm) = 5.39 eV
E3  E2 transition
E9  E3 transition
E3= hc/ (1243 eVnm)/(541 nm) = 2.30 eV
E7  E4 transition
10. A hydrogen atom is in its tenth excited state according to the Borh model (n = 11).
a. What is the radius of the Bohr orbit?
rn = n2a0
r11 = (11)2(0.0529 nm) = 6.40 nm
b. What is the angular momentum of the electron?
Ln = n h/(2)
L11 =1 h/(2) = 1.16 x 10-33 J-s
c. What is the electron’s kinetic energy?
L=pr
p = L/r = 1.78 x 10-25 kg-m/s
KE = p2/(2me) = (1.78 x 10-25 kg-m/s)2/(2(9.11 x 10-31 kg)) = 1.73 x 10-20 J = 0.112 eV
d. What is the electron’s potential energy?
PE = - ke e2/r
PE = - (8.99 x 109 N-m2/kg2) (1.60 x 10-19 C)2/(6.40 x 10-9 m) = -0.224 eV
e. What is the electron’s total energy?
E = KE + PE = 0.112 eV + (-0.224 eV) = -0.112 eV
TIME
OF
COMPLETION_______________
NAME____SOLUTION_________________________
DEPARTMENT OF NATURAL SCIENCES
PHYS 3650, Exam 2
Version 1
Total Weight: 100 points
Section 1
November 4, 2006
1.
Check your examination for completeness prior to starting. There are a total of ten (10)
problems on seven (7) pages.
2.
Authorized references include your calculator with calculator handbook, and the Reference
Data Pamphlet (provided by your instructor).
3.
You will have 50 minutes to complete the examination.
4.
The total weight of the examination is 100 points.
5.
There are six (6) multiple choice and four (4) calculation problems. Work five (5) multiple
choice and three (3) calculation problems. Show all work; partial credit will be given for correct
work shown.
6.
If you have any questions during the examination, see your instructor who will
be located in the classroom.
7.
Start:
Stop:
11:30 a.m.
12:20 p.m
PROBLEM
POINTS
1-6
25
7
25
8
25
9
25
10
25
TOTAL
100
CREDIT
PERCENTAGE
CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN
MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR
PARTIAL CREDIT.
1. According to the Bohr model of the atom, the angular momentum of an electron around the
nucleus
a. Could equal any positive value.
b. Must equal an integral multiple of h.
(5)
c. Must equal an integral multiple of h/2π.
d. Decreases with time, eventually becoming zero.
2. The reason the position of a particle cannot be specified with infinite precision is the
a. Exclusion principle.
b. Uncertainty principle.
(5)
c. Photoelectric effect.
d. Principle of relativity.
3. The principal quantum number can have any integer value ranging from
a. -∞ to +∞.
b. 0 to ∞.
(5)
c. 1 to ∞.
d. 1 to 100.
4. The spin quantum number can have values of
a. -1/2, -1, 0, +1, +1/2.
b. -1/2, -1, +1, +1/2.
(5)
c. -1/2, 0, +1/2.
d. -1/2, +1/2.
5. In the ground state, the quantum numbers (n, l, ml, ms) for hydrogen are, respectively,
a. 1, 1, 1, 1.
b. 1, 0, 0, 0.
(5)
c. 1, 0, 0, ±1/2.
d. 1, 1, 1, ±1/2.
6. In terms of an atom's electron configuration, the letters K, L, M, and N refer to
a. Different shells with n equal to 1, 2, 3, or 4 respectively.
b. Different sub shells with l equal to 1, 2, 3, or 4 respectively.
(5)
c. The four possible levels for the magnetic quantum number.
d. The four possible quantum numbers.
7. What is the full electron configuration in the ground state for elements with Z equal to
(a) 10,
1s22s22p6
(b) 16,
1s22s22p63s23p4
(c) 28?
1s22s22p63s23p64s23d8
8. Use the Bohr theory to estimate the wavelength for an n  3 to n  1 transition in molybdenum
(Z  42). The measured value is 0.063 nm. Why do we not expect perfect agreement?
EK = - (Z-1)2 E0/12 = -22,862 eV
EM = - (Z-9)2 E0/32 = -1,64 eV
E = EM – EK = 21,216 eV
 = hc/E = 0.0584 nm
9. (a) Find the photon energy and wavelength for the series limit (shortest wavelength) in the
Paschen series (nf = 3).
1/ = R(1/nf2 – 1/ni2) = (1.10 x 107 m-1)(1/32 – 1/inf2) = 0.12222 x 107 m-1
 = 8.18 x 10-7 m = 818 nm
(b) Calculate the wavelengths for the three longest wavelengths in this series.
1/ = R(1/nf2 – 1/ni2) = (1.10 x 107 m-1)(1/32 – 1/42) = 0.05347 x 107 m-1
 = 18.70 x 10-7 m = 1870 nm
1/ = R(1/nf2 – 1/ni2) = (1.10 x 107 m-1)(1/32 – 1/52) = 0.07111 x 107 m-1
 = 14.06 x 10-7 m = 1406 nm
1/ = R(1/nf2 – 1/ni2) = (1.10 x 107 m-1)(1/32 – 1/62) = 0.08333 x 107 m-1
 = 12.00 x 10-7 m = 1200 nm
10. Make a table of all of the allowed four quantum numbers for the first three shells of the
hydrogen atom. How many electrons can each shell accommodate?
K (2 electrons): n = 1
l=0
(1 0 0 -½ ) 2 1s electrons
ml = 0
ms = +- 1/2
(1 0 0 ½ )
L (8 electrons): n = 2
l=0
(2 0 0 -½ ) 2 2s electrons
ml = 0
ms = +- 1/2
(2 0 0 ½ )
l=1
ml = -1
ms = +- 1/2
(2 1 -1 ½ )
(2 1 -1 -½ )
1 0 -½ )
l=1
6 2p electrons
ml = 0
ms = +- 1/2
(2 1 0 ½ )
(2
l=1
ml = 1
ms = +- 1/2
(2 1 1 ½ )
(2
1 1 -½ )
M (18 electrons): n = 3
l=0
(3 0 0 -½ ) 2 3s electrons
ml = 0
ms = +- 1/2
(3 0 0 ½ )
l=1
ml = -1
ms = +- 1/2
(3 1 -1 ½ )
(3 1 -1 -½ )
(3 1 0 -½ )
l=1
6 3p electrons
ml = 0
ms = +- 1/2
(3 1 0 ½ )
l=1
ml = 1
ms = +- 1/2
(3 1 1 ½ )
l=2
ml = -2
(3 1 1 -½ )
(3 2 -2 -½ )
ms = +- 1/2
(3 2 -2 ½ )
l=2
ml = -1
ms = +- 1/2
(3 2 -1 ½ )
(3 2 -1 -½ )
(3 2 0 -½ )
l=2
10 3d electrons
ml = 0
ms = +- 1/2
(3 2 0 ½ )
l=2
ml = 1
ms = +- 1/2
(3 2 1 ½ )
l=2
ml = 2
ms = +- 1/2
(3 2 2 ½ )
(3 2 1 -½ )
(3 2 2 -½ )
TIME
OF
COMPLETION_______________
NAME____SOLUTION_________________________
DEPARTMENT OF NATURAL SCIENCES
PHYS 3650, Exam 2
Version 1
Total Weight: 100 points
Section 1
October 31, 2008
1.
Check your examination for completeness prior to starting. There are a total of nine (9)
problems on six (6) pages.
2.
Authorized references include your calculator with calculator handbook, and the Reference
Data Pamphlet (provided by your instructor).
3.
You will have 50 minutes to complete the examination.
4.
The total weight of the examination is 100 points.
5.
There are six (6) multiple choice and three (3) calculation problems. Work five (5) multiple
choice and three (3) calculation problems. Show all work; partial credit will be given for correct
work shown.
6.
If you have any questions during the examination, see your instructor who will
be located in the classroom.
7.
Start:
Stop:
PROBLEM
10:00 a.m.
10:50 a.m
POINTS
CREDIT
1-6
25
7
25
8
25
9
25
TOTAL
100
PERCENTAGE
CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN
MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR
PARTIAL CREDIT.
1. Hydrogen atoms can emit lines with visible colors from red to violet. These four visible lines
emitted by hydrogen atom are produced by electrons
a. That start in the n = 2 level.
b. That end up the n = 2 level.
(5)
c. That end up in the n = 3 level.
d. That start in the ground level.
2. According to the Pauli’s exclusion principle, how many electrons in an atom may have a
particular set of quantum numbers?
a. 1.
b. 2.
(5)
c. 6.
d. 10.
3. The orbital angular momentum quantum number can take which of the following values for
any given value of the principal quantum number, n?
a. l = 0, 1, 2,…..
b. l = 0, 1, 2, …., n.
(5)
c. l = 0, 1, 2, …., n – 1.
d. l = 0, 1, 2, …., n + 1.
4. Which of the following values are associated with the electron spin quantum number, ms?
a. ± 1/2.
b. 0.
(5)
c. ± 1.
d. ± 2.
5. Which ones of the atomic transition on the sodium energy level diagram below are NOT
allowed by the selection rules?
a. a and c.
b. b and f.
(5)
c. c and d.
d. d and g.
6. Given the energy diagram above, estimate the wavelength of a photon emitted by a sodium
atom as the electron goes through the transition d.
a. 800 nm.
(5)
b. 620 nm.
c. 450 nm.
d. 75 nm.
7. Titanium (Z = 22) is in its ground state.
a. What is its electronic configuration?
1s22s22p63s23p64s23d2
b. List all the possible combinations of quantum numbers (n, l, ml, and ms)
an electron can have while in the subshell which is only partially filled in the ground
state.
3d subshell: n = 3, l = 2
l=2
ml = -2
ms = +- 1/2
(3 2 -2 ½ )
l=2
ml = -1
ms = +- 1/2
(3 2 -1 ½ )
l=2
ml = 0
ms = +- 1/2
(3 2 0 ½ )
l=2
ml = 1
ms = +- 1/2
(3 2 1 ½ )
l=2
ml = 2
ms = +- 1/2
(3 2 2 ½ )
(3 2 -2 -½ )
(3 2 -1 -½ )
(3 2 0 -½ )
(3 2 1 -½ )
(3 2 2 -½ )
8. Using Bohr’s model, estimate the wavelength of the K line for
a. calcium (Z = 20), and
EK = - (Z-1)2 E0/12 = - (19)2 (13.6 eV)/12 = - 4,910 eV
EL = - (Z-1)2 E0/12 = - (19)2 (13.6 eV)/22 = - 1.227 eV
E = EL – EK = 3,683 eV
 = hc/E = 0.337 nm
b. cadmium (Z = 48)
EK = - (Z-1)2 E0/12 = - (47)2 (13.6 eV)/12 = - 30,042 eV
EL = - (Z-1)2 E0/12 = - (47)2 (13.6 eV)/22 = - 7,510 eV
E = EL – EK = 22,532 eV
 = hc/E = 0.0550 nm
10. Find the energy of the electron in the ground state of singly ionized helium (Z = 2).
E1 = - (Z)2 E0/12 = - (2)2 (13.6 eV)/12 = - 54.4 eV
What is the shortest possible wavelength which is emitted as a result of electron returning to
the ground state?
Shortest wavelength corresponds to the transition from n = ∞:
E = E∞ – E1 = 54.4 eV
 = hc/E = 22.8 nm
What is the longest possible wavelength which is emitted as a result of electron returning to
the ground state?
Lonest wavelength corresponds to the transision from n = 2:
E2 = - (Z)2 E0/22 = - (2)2 (13.6 eV)/22 = - 13.6 eV
E = E2 – E1 = 40.8 eV
 = hc/E = 30.4 nm
TIME
OF
COMPLETION_______________
NAME__SOLUTION___________________________
DEPARTMENT OF NATURAL SCIENCES
PHYS 3650, Exam 2
Version 1
Total Weight: 100 points
Section 1
November 13, 2009
1.
Check your examination for completeness prior to starting. There are a total of nine (9)
problems on six (6) pages.
2.
Authorized references include your calculator with calculator handbook, and the Reference
Data Pamphlet (provided by your instructor).
3.
You will have 50 minutes to complete the examination.
4.
The total weight of the examination is 100 points.
5.
There are six (6) multiple choice and three (3) calculation problems. Work five (5) multiple
choice and three (3) calculation problems. Show all work; partial credit will be given for correct
work shown.
6.
If you have any questions during the examination, see your instructor who will
be located in the classroom.
7.
Start:
Stop:
10:00 a.m.
10:50 a.m
PROBLEM
POINTS
1-6
25
7
25
8
25
CREDIT
9
25
TOTAL
100
PERCENTAGE
CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN
MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR
PARTIAL CREDIT.
1. The Pauli exclusion principle:
a. Implies that in an atom no two electrons can have the same set of quantum numbers.
b. Says that no two electrons in an atom can have the same orbit.
(5)
c. Excludes electrons from atomic nuclei.
d. Excludes protons from atomic orbits.
2. Which one of these statements is true?
a. The principal quantum number of the electron in a hydrogen atom does not affect its
energy.
b. The principal quantum number of an electron in its ground state is zero.
(5)
c. The orbital quantum number of an electron state is always less than the principal
quantum number of that state.
d. The electron spin quantum number can take on any one of the four different values.
3. The orbital angular momentum quantum number can take which of the following values for
any given value of the principal quantum number, n?
a. l = 0, 1, 2,…..
b. l = 0, 1, 2, …., n.
(5)
c. l = 0, 1, 2, …., n – 1.
d. l = 0, 1, 2, …., n + 1.
4. How many of oxygen’s eight electrons are found in the p state?
a. 0.
b. 2.
(5)
c. 4.
d. 6.
5. The total number of states of hydrogen with principal quantum number n = 4 is:
a. 4.
b. 16.
(5)
c. 32.
d. 36.
6. Given the sodium energy diagram above, estimate the wavelength of a photon emitted by a
PHYS 3650 Exam 2, Version 1
Fall 2005
19
sodium atom as the electron goes through the transition d.
a. 800 nm.
(5)
b. 620 nm.
c. 450 nm.
d. 75 nm.
9. Nickel (Z = 28) is in its ground state.
a. What is its electronic configuration?
1s22s22p63s23p64s23d8
b. List all the possible combinations of quantum numbers (n, l, ml, and ms)
an electron can have while in the subshell which is only partially filled in the ground
state.
n = 3; l = 2; ml = -2, -1, 0, 1, 2; ms = ±1/2
(3, 2, -2, ±1/2) (3, 2, -1, ±1/2) (3, 2, 0, ±1/2) (3, 2, 1, ±1/2) (3, 2, 2, ±1/2)
10. Using Bohr’s model, estimate the wavelength of the K line for a nickel target (Z = 28).
EK = - (Z-1)2 E0/12 = - (28-1)2 (13.6 eV)/12 = - 9,914 eV
EL = - (Z-1)2 E0/22 = - (28-1)2 (13.6 eV)/22 = - 2.479 eV
E = EL – EK = 7,435 eV
 = hc/E = 0.167 nm
What is the wavelength of the K line?
PHYS 3650 Exam 2, Version 1
Fall 2005
20
EK = - (Z-1)2 E0/12 = - (28-1)2 (13.6 eV)/12 = - 9,914 eV
EM = - (Z-9)2 E0/32 = - (28-9)2 (13.6 eV)/32 = - 546 eV
E = EM – EK = 9,368 eV
 = hc/E = 0.132 nm
11. Find the energy of the electron in the ground state of doubly ionized lithium (Z = 3).
E1 = - (Z)2 E0/12 = - (3)2 (13.6 eV)/12 = - 122 eV
What is the shortest possible wavelength which is emitted as a result of electron returning to
the ground state?
E = 0 – E1 = 122 eV
 = hc/E = 10.1 nm
What is the longest possible wavelength which is emitted as a result of electron returning to
the ground state?
E2 = - (Z)2 E0/22 = - (3)2 (13.6 eV)/22 = - 30.6 eV
E = E2 – E1 = 91.8 eV
 = hc/E = 13.5 nm
TIME
OF
COMPLETION_______________
NAME____SOLUTION_________________________
DEPARTMENT OF NATURAL SCIENCES
PHYS 3650, Exam 2
PHYS 3650 Exam 2, Version 1
Fall 2005
Section 1
21
Version 1
Total Weight: 100 points
March 28, 2011
1.
Check your examination for completeness prior to starting. There are a total of nine (9)
problems on six (6) pages.
2.
Authorized references include your calculator with calculator handbook, and the Reference
Data Pamphlet (provided by your instructor).
3.
You will have 50 minutes to complete the examination.
4.
The total weight of the examination is 100 points.
5.
There are six (6) multiple choice and three (3) calculation problems. Work all calculation
problems and 5 (five) multiple choice. Show all work; partial credit will be given for correct work
shown.
6.
If you have any questions during the examination, see your instructor who will
be located in the classroom.
7.
Start:
Stop:
3:00 p.m.
3:50 p.m.
PROBLEM
POINTS
1-6
25
7
25
8
25
9
25
TOTAL
100
CREDIT
PERCENTAGE
CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN
MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR
PARTIAL CREDIT.
PHYS 3650 Exam 2, Version 1
Fall 2005
22
1. In a Compton scattering experiment, a photon of energy E is scattered from an electron at
rest. After the scattering event occurs, which of the following statements is true?
(A)
The frequency of the photon is greater than E/h.
(B)
The energy of the photon is less than E.
(C)
The wavelength of the photon is less than hc/E.
(D)
The momentum of the photon increases.
(E)
None of those statements is true.
(5)
2. Which of the following phenomena most clearly demonstrates the wave nature of electrons?
(A) The photoelectric effect.
(B) The blackbody radiation.
(5)
(C) The Compton effect.
(D) Diffraction of electrons by crystals.
(E) None of these answers.
3. A proton, an electron, and a helium nucleus all move at speed v. Rank their de Broglie
wavelengths from largest to smallest.
(A)
electron, proton, helium nucleus.
(B)
proton, helium nucleus, electron.
(C)
proton, electron, helium nucleus.
(D)
helium nucleus, electron, proton.
(E)
helium nucleus, proton, electron
(5)
PHYS 3650 Exam 2, Version 1
Fall 2005
23
4.What is the longest wavelength in the Lyman Series?
(A)
45.60 nm.
(B)
91.20 nm.
(C)
121.5 nm.
(D)
240.1 nm.
(E)
356.2 nm.
5. To which of the following values of n does the longest wavelength in the Balmer series
correspond?
(A) 1.
(B) 3.
(5)
(C) 5.
(D) 7.
(E) ∞ .
6. Which one of the following is the correct expression for the energy of a photon?
(A)
E = h/f.
(B) E = h/c.
(5)
(C) E = h.
(D) E = hc/.
(F)
None of these.
7. The work function for aluminum is 4.08 eV.
(a)
Find the cutoff wavelength for aluminum.
PHYS 3650 Exam 2, Version 1
Fall 2005
24
0 
hc


1240 eV  nm
 304 nm
4.08 eV
(b) What is the lowest frequency of light incident on aluminum that releases
photoelectrons from its surface?
f 0
c
0

3.00 108 m / s
 9.87 1014 Hz
304 nm
(c) If photons of energy 5.81 eV are incident on aluminum, what is the maximum kinetic
energy of the ejected photoelectrons?
KE  hf    5.81 eV  4.08 eV  1.73 eV
8. In the Compton effect, a 0.100-nm photon strikes a free electron in a head-on collision and
knocks it into the forward direction. The rebounding photon recoils directly backward. Find
a. The wavelength of the scattered photon.
 
h
(1  cos  )  (0.00243 nm)(1  cos(180o ))  0.00486 nm
me c
    0.100 nm  0.00486 nm  0.10486 nm
0
b. The energy of the scattered photon.
E
hc


1240 eV  nm
 11,825 eV
0.10486 nm
c. The kinetic energy of the recoiling electron. (Hint: subtract the final
energy of the photon from its initial energy to find the kinetic energy of
the electron.)
E0 
hc
0

1240 eV  nm
 12,400 eV
0.100 nm
PHYS 3650 Exam 2, Version 1
Fall 2005
25
KE  E0  E  0.600 keV
9. (a) Find the photon energy and wavelength for the series limit (shortest wavelength) in the
Paschen series (n’= 3).
1

R H (
R
1
1
 2) H
2
9
3 
  818 nm
E
hc


1240 eV  nm
 1.52 eV
818 nm
(b)
1

R H (
Calculate the wavelengths for the three longest wavelengths in this series.
1 1
 )
32 4 2
  1870 nm
1

R H (
1 1
 )
32 5 2
  1278 nm
1

R H (
1 1
 )
32 6 2
  1091 nm
PHYS 3650 Exam 2, Version 1
Fall 2005
26

PHYS 3650 Exam 2, Version 1
Fall 2005
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PHYS 3650 Exam 2, Version 1
Fall 2005
28