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tiL12214_appe_643-698.indd Page 664 09/10/10 8:37 AM user-f463 Volume/207/MHDQ243/tiL12214_disk1of1/0073512214/tiL12214_pagefile

–20
8.4. For n = 6, E H = –6.05 × 10 J
CHAPTER 8
–19
For n = 2, E L = –5.44 × 10 J
–27
8.1. m = 1.68 × 10 kg –34
h = 6.63 × 10 J·s
3
v = 3.22 × 10 m/s
f = ?
–34
h = 6.63 × 10 J·s
E H – E L

hf = E H – E L ∴ f = _

λ = ? h
–19
h _ (–6.05 × 10 J) – (–5.44 × 10 J)
–20

λ = ___

mv f =
–34
6.63 × 10 J·s
–34
6.63 × 10 J·s
___
–19
J
= 4.84 × 10 _
__

3 m _

–27

(1.68 × 10 kg) ( 3.22 × 10 ) = –34 J·s
s 6.63 × 10
–34
J·s
____ 14 1 _
6.63 × 10
= = 7.29 × 10

m _
s
–27
3
(1.68 × 10 )(3.22 × 10 ) kg ×

s 14
= 7.29 × 10 Hz
kg·m
_ 2
× s
–34

6.63 × 10 _ E 1 _
__
s·s

= m _ 8.5. (n = 1) = –13.6 eV E n =
5.41 × 10 –24 kg × n 2

s E = ?
_

kg·m·m
–10 _ _ _ = –13.6 eV
s
1

= 1.23 × 10 1 2
× ×
s kg m
= –13.6 eV
–10
= 1.23 × 10 m
Since the energy of the electron is –13.6 eV, it will require
8.2. (a) n = 6 E L _ –18
E n = 13.6 eV (or 2.17 × 10 J) to remove the electron.

E L = –13.6 eV n 2
11
8.6. q/m = –1.76 × 10 C/kg __
charge
–13.6 eV
E 6 = ? _ mass =

E 6 = –19 charge/mass
6 2 q = –1.60 × 10 C
–19
–1.60 × 10 C

_ m = ? = __
–13.6 eV

= 11 C _
36 –1.76 × 10

kg
= –0.378 eV
–19
__ kg
_
–1.60 × 10

J
–19 _ = C ×
(b) = (–0.378 eV)( 1.60 × 10 ) –1.76 × 10 –11 C

eV –31
J
_
–19

= (–0.378)(1.60 × 10 ) eV × = 9.09 × 10 kg
eV 8.7. λ = 1.67 × 10 m
–10
–20
= –6.05 × 10 J m = 9.11 × 10 kg
–31
8.3. (a) Energy is related to the frequency and Planck’s constant in v = ?
equation 8.1, E = hf. From equation 8.4, h _ h _

λ = ∴ v =
mλ
mv
hf = E H – E L ∴ E = E H – E L
–34
6.63 × 10 J·s
–20
For n = 6, E H = 6.05 × 10 J v = ___

–10
–31
–19
For n = 2, E L = 5.44 × 10 J (9.11 × 10 kg) (1.67 × 10 m)
–34
___ _
J·s
6.63 × 10
E = ? J =

–10
–31
(9.11 × 10 ) (1.67 × 10 ) kg·m
E = E H – E L 2
kg·m
_
–19
–20

= (–6.05 × 10 J) – (–5.44 × 10 J) ___

× s

–34
s·s
6.63 × 10

–19
= 4.84 × 10 J = –40
1.52 × 10 kg·m
(b) E H = –0.377 eV* 6 _ _ _
kg·m·m
1
1

= 4.36 × 10
× ×
E L = –3.40 eV* s kg m
E = ? eV 6 m _
= 4.36 × 10

s
E = E H – E L
2
2
= (–0.377 eV) – (–3.40 eV) 8.8. (a) Boron: 1s 2s 2p 1
6
2
2
2
(b) Aluminum: 1s 2s 2p 3s 3p 1
= 3.02 eV 2 2 6 2 6 1
(c) Potassium: 1s 2s 2p 3s 3p 4s
*From Figure 8.11.
8.9. (a) Boron is atomic number 5 and there are 5 electrons.
(b) Aluminum is atomic number 13 and there are 13 electrons.
(c) Potassium is atomic number 19 and there are 19 electrons.
664 APPENDIX E Solutions for Group A Parallel Exercises E-22