Practice Problems (Chapter 7) Quantum Theory and Atomic Structure - KEY
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Practice Problems (Chapter 7): Quantum Theory CHEM 1A 1. Hydrogen has a red emission line at 656.3 nm, what is the energy and frequency of a photon of this light? Note: Planck’s constant = 6.626 x 10 –34 J·s, speed of light = 2.998 x 10 8 m/s 2. What is the de Broglie wavelength for an electron moving at 2.42 x 10 6 m/s? Note: m e = 9.109 x 10 –31 kg 3. Using the Bohr equation, calculate is the change in energy when an electron transitions from n = 3 to n= 2 in a hydrogen atom. KEY Answer (E): ___________________ 3.027 x 10 –19 J Answer (): ___________________ E photon = 3 .026 77861 x 10 –19 J E photon = h = hc E photon = (6.626 x 10 –34 J·s) (2.998 x 10 8 m/s) 656.3 nm 10 –9 m 1 nm = 4 .568 032912 x 10 14 s –1 = c = (2.998 x 10 8 m/s) 656.3 nm 10 –9 m 1 nm 4.568 x 10 14 s –1 or Hz 1 2 2 2 1 2 3 2 E = –2.18 x 10 –18 J – E = – 3 .02 777778 x 10 –19 J Answer: ___________________ – 3.03 x 10 –19 J = h mu = 6.626 x 10 –34 J·s (9.109 x 10 –31 kg) (2.42 x 10 6 m/s) = 3 .00 583657 x 10 –10 m Answer: ___________________ 3.01 x 10 –10 m many orders of magnitude larger than the diameter of an electron Z 2 n Z 2 n E = –2.18 x 10 –18 J –
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Transcript of Practice Problems (Chapter 7) Quantum Theory and Atomic Structure - KEY
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Practice Problems (Chapter 7): Quantum Theory
CHEM 1A
1. Hydrogen has a red emission line at 656.3 nm, what is the energy and frequency of a photon of this light? Note: Plancks constant = 6.626 x 1034 Js, speed of light = 2.998 x 108 m/s
2. What is the de Broglie wavelength for an electron moving at 2.42 x 106 m/s? Note: me = 9.109 x 10
31 kg
3. Using the Bohr equation, calculate is the change in energy when an electron transitions from n = 3 to n= 2 in a hydrogen atom.
KEY
Answer (E): ___________________ 3.027 x 1019
J Answer (): ___________________
Ephoton = 3.02677861 x 1019
J
Ephoton = h = hc
Ephoton = (6.626 x 10
34 Js) (2.998 x 10
8 m/s)
656.3 nm 10
9 m
1 nm
= 4.568032912 x 1014 s1
= c
= (2.998 x 10
8 m/s)
656.3 nm 10
9 m
1 nm
4.568 x 1014
s1
or Hz
12
22
12
32
E = 2.18 x 1018 J
E = 3.02777778 x 1019 J
Answer: ___________________ 3.03 x 1019
J
= h
mu
= 6.626 x 10
34 Js
(9.109 x 1031
kg) (2.42 x 106 m/s)
= 3.00583657 x 1010
m
Answer: ___________________ 3.01 x 1010
m many orders of magnitude larger than the diameter of an electron
Z2
n
Z2
n
E = 2.18 x 1018 J
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4. Using the Rydberg formula, calculate the initial energy level when an electron in a hydrogen atom transitions into n= 2 and emits a photon at 410.1 nm.
Note: the Rydberg constant = 1.097 x 107 m
1
5. What is the maximum number of electrons that are allowed to have the following set of quantum numbers in one atom?
n = 4 and ml = +2 Answer: ______
n = 3 and l = 1 Answer: ______
n = 1 and ms = +1/2 Answer: ______
6. Provide the possible values for the other three quantum numbers for electrons in n = 3.
7. What are the four quantum numbers for the last electron in each of the following elements? Note: Orbital notation (Chapter 8) makes this easier. Assume that ml values fill from more negative to
more positive, and that a spin up e (ms = +
1/2) each fill an orbital before spin down e
begin to pair.
Possible Values
l 0 1 2
ml 0 1, 0, +1 2, 1, 0, +1, +2
ms +1/2,
1/2 +
1/2,
1/2 +
1/2,
1/2
n l ml ms
nitrogen 2 1 +1 +1/2
sulfur 3 1 1 1/2
copper 3 2 +2 1/2
2p 2s
3p 3s
3d 4s
Answer: ___________________ n = 6
4 e
only l = 2 (d) and l = 3 (f) on n = 4 can have
ml = +2 orbitals, each of which can hold 2 e
6 e
l = 1 (p) has 3 orbitals (ml = 1, 0, +1), each of which can hold 2 e
1 e
n = 1 has 1 sublevel (l = 0 (s)) with 1 orbital
(ml = 0), which can hold 1 spin up e
Max. # e: ______ 18 e
Z2
n
Z2
n
= R
1
ninit = 6.006428031
uncertain digit
12
22
12
n
= 1.097 x 10
7 m
1
410.1 nm 10
9 m
1 nm
1
