time of completion - clayton state university · time of completion_____ name ... de broglie. (5)...

TIME OF COMPLETION_______________

NAME______SOLUTION_______________________

DEPARTMENT OF NATURAL SCIENCES

PHYS 3650, Exam 2 Section 1

Version 1 October 31, 2005

Total Weight: 100 points

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: 10:30 a.m.

Stop: 11:45 a.m

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

20

9

15

10

15

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 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: 1s22s

1

Be: 1s22s

2

Na: 1s22s

22p

63s

1

Mg: 1s22s

22p

63s

2

K: 1s22s

22p

63s

23p

64s

1

Ca: 1s22s

22p

63s

23p

64s

2

Cr: 1s22s

22p

63s

23p

64s

23d

4

Ni: 1s22s

22p

63s

23p

64s

23d

8

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 E0/1

2 = -26,330 eV

EL = - (Z-1)2 E0/2

2 = -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 E0/3

2 = -1,958 eV

EL = - (Z-1)2 E0/2

2 = -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/n

2

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 E3 E2 transition

E2= hc/ (1243 eVnm)/(230.6 nm) = 5.39 eV 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 = p r

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-m

2/kg

2) (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


NAME____SOLUTION_________________________



Version 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.





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.



7. Start: 11:30 a.m.

Stop: 12:20 p.m

PROBLEM

POINTS

CREDIT

1-6

25

7

25

8

25

9

25

10

25

TOTAL

100

PERCENTAGE



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,

1s22s

22p

6

(b) 16,

1s22s

22p

63s

23p

4

(c) 28?

1s22s

22p

63s

23p

64s

23d

8

8. Use the Bohr theory to estimate the wavelength for an 3n to 1n transition in molybdenum

).42( Z The measured value is 0.063 nm. Why do we not expect perfect agreement?

EK = - (Z-1)2 E0/1

2 = -22,862 eV

EM = - (Z-9)2 E0/3

2 = -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/ni

2) = (1.10 x 10

7 m

-1)(1/3

2 – 1/inf

2) = 0.12222 x 10

7 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/ni

2) = (1.10 x 10

7 m

-1)(1/3

2 – 1/4

2) = 0.05347 x 10

7 m

-1

= 18.70 x 10-7

m = 1870 nm

1/ = R(1/nf2 – 1/ni

2) = (1.10 x 10

7 m

-1)(1/3

2 – 1/5

2) = 0.07111 x 10

7 m

-1

= 14.06 x 10-7

m = 1406 nm

1/ = R(1/nf2 – 1/ni

2) = (1.10 x 10

7 m

-1)(1/3

2 – 1/6

2) = 0.08333 x 10

7 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 ml = 0 ms = +- 1/2 (1 0 0 ½ )

(1 0 0 -½ ) 2 1s electrons

L (8 electrons): n = 2 l = 0 ml = 0 ms = +- 1/2 (2 0 0 ½ )

(2 0 0 -½ ) 2 2s electrons

l = 1 ml = -1 ms = +- 1/2 (2 1 -1 ½ )

(2 1 -1 -½ )

l = 1 ml = 0 ms = +- 1/2 (2 1 0 ½ ) (2

1 0 -½ ) 6 2p electrons

l = 1 ml = 1 ms = +- 1/2 (2 1 1 ½ ) (2

1 1 -½ )

M (18 electrons): n = 3 l = 0 ml = 0 ms = +- 1/2 (3 0 0 ½ )

(3 0 0 -½ ) 2 3s electrons

l = 1 ml = -1 ms = +- 1/2 (3 1 -1 ½ )

(3 1 -1 -½ )

l = 1 ml = 0 ms = +- 1/2 (3 1 0 ½ )

(3 1 0 -½ ) 6 3p electrons

l = 1 ml = 1 ms = +- 1/2 (3 1 1 ½ )

(3 1 1 -½ )

l = 2 ml = -2 ms = +- 1/2 (3 2 -2 ½ )

(3 2 -2 -½ )

l = 2 ml = -1 ms = +- 1/2 (3 2 -1 ½ )

(3 2 -1 -½ )

l = 2 ml = 0 ms = +- 1/2 (3 2 0 ½ )

(3 2 0 -½ ) 10 3d electrons

l = 2 ml = 1 ms = +- 1/2 (3 2 1 ½ )

(3 2 1 -½ )

l = 2 ml = 2 ms = +- 1/2 (3 2 2 ½ )

(3 2 2 -½ )


NAME____SOLUTION_________________________



Version 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.





5. There are six (6) multiple choice and three (3) calculation problems. Work five (5) multiple


work shown.



7. Start: 10:00 a.m.

Stop: 10:50 a.m

PROBLEM

POINTS

CREDIT

1-6

25

7

25

8

25

9

25

TOTAL

100

PERCENTAGE



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?

1s22s

22p

63s

23p

64s

23d

2

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 )

(3 2 -2 -½ )

l = 2 ml = -1 ms = +- 1/2 (3 2 -1 ½ )

(3 2 -1 -½ )

l = 2 ml = 0 ms = +- 1/2 (3 2 0 ½ )

(3 2 0 -½ )

l = 2 ml = 1 ms = +- 1/2 (3 2 1 ½ )

(3 2 1 -½ )

l = 2 ml = 2 ms = +- 1/2 (3 2 2 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 E0/1

2 = - (19)

2 (13.6 eV)/1

2 = - 4,910 eV

EL = - (Z-1)2 E0/1

2 = - (19)

2 (13.6 eV)/2

2 = - 1.227 eV

E = EL – EK = 3,683 eV

= hc/E = 0.337 nm

b. cadmium (Z = 48)

EK = - (Z-1)2 E0/1

2 = - (47)

2 (13.6 eV)/1

2 = - 30,042 eV

EL = - (Z-1)2 E0/1

2 = - (47)

2 (13.6 eV)/2

2 = - 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 E0/1

2 = - (2)

2 (13.6 eV)/1

2 = - 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 E0/2

2 = - (2)

2 (13.6 eV)/2

2 = - 13.6 eV

E = E2 – E1 = 40.8 eV

= hc/E = 30.4 nm


NAME__SOLUTION___________________________



Version 1 November 13, 2009








5. There are six (6) multiple choice and three (3) calculation problems. Work five (5) multiple


work shown.



7. Start: 10:00 a.m.

Stop: 10:50 a.m

PROBLEM

POINTS

CREDIT

1-6

25

7

25

8

25

9 25 TOTAL

100

PERCENTAGE



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,…..

PHYS 3650 Exam 2, Version 1

Fall 2005 19

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


Fall 2005 20

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?

1s22s

22p

63s

23p

64s

23d

8

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 E0/1

2 = - (28-1)

2 (13.6 eV)/1

2 = - 9,914 eV

EL = - (Z-1)2 E0/2

2 = - (28-1)

2 (13.6 eV)/2

2 = - 2.479 eV

E = EL – EK = 7,435 eV

= hc/E = 0.167 nm

What is the wavelength of the K line?


Fall 2005 21

EK = - (Z-1)2 E0/1

2 = - (28-1)

2 (13.6 eV)/1

2 = - 9,914 eV

EM = - (Z-9)2 E0/3

2 = - (28-9)

2 (13.6 eV)/3

2 = - 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 E0/1

2 = - (3)

2 (13.6 eV)/1

2 = - 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 E0/2

2 = - (3)

2 (13.6 eV)/2

2 = - 30.6 eV

E = E2 – E1 = 91.8 eV

= hc/E = 13.5 nm


NAME____SOLUTION_________________________




Fall 2005 22

Version 1 March 28, 2011








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.



7. Start: 3:00 p.m.

Stop: 3:50 p.m.

PROBLEM

POINTS

CREDIT

1-6

25

7

25

8

25

9

25

TOTAL

100

PERCENTAGE



PARTIAL CREDIT.


Fall 2005 23

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.

(5)

(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.

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.

(5)

(C) proton, electron, helium nucleus.

(D) helium nucleus, electron, proton.

(E) helium nucleus, proton, electron


Fall 2005 24

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.


Fall 2005 25

nmeV

nmeVhc304

08.4

12400

(b) What is the lowest frequency of light incident on aluminum that releases

photoelectrons from its surface?

Hznm

smcf 14

8

0

0 1087.9304

/1000.3

(c) If photons of energy 5.81 eV are incident on aluminum, what is the maximum kinetic

energy of the ejected photoelectrons?

eVeVeVhfKE 73.108.481.5

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.

nmnmcm

h o

e

00486.0))180cos(1)(00243.0()cos1(

nmnmnm 10486.000486.0100.00

b. The energy of the scattered photon.

eVnm

nmeVhcE 825,11

10486.0

1240

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.)

eVnm

nmeVhcE 400,12

100.0

1240

0

0


Fall 2005 26

keVEEKE 600.00

9. (a) Find the photon energy and wavelength for the series limit (shortest wavelength) in the

Paschen series (n’= 3).

9)

1

3

1(

122

HH

RR

nm818

eVnm

nmeVhcE 52.1

818

1240

(b) Calculate the wavelengths for the three longest wavelengths in this series.

)4

1

3

1(

122

HR

nm1870

)5

1

3

1(

122

HR

nm1278

)6

1

3

1(

122

HR

nm1091


Fall 2005 27


Fall 2005 28

time of completion - clayton state university · time of completion_____ name ... de broglie. (5)...

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