(b)
− 2.0 eV = 2.97 eV = 4.75 × 10
−19 J
(c)
− 2.0 eV = 1.55 eV = 2.47 × 10
−19 J
9.14 (a)
− 2.3 eV = 3.88 eV = 6.2 × 10
−19 J
(b)
− 2.3 eV = 2.67 eV = 4.27 × 10
−19 J
(c)
− 2.3 eV = 1.25 eV = 2.0 × 10
−19 J
9.15 (a)
(b)
(c)
9.16 The minimal frequency corresponds to a minimal energy (hν) needed
to overcome the work function.
9.17
9.18 According to classical theory, the energy is independent of frequency,
but according to quantum theory, the energy is proportional to the
frequency (E = hν).
9.19 4m
3
ν = 4m
2
p → 4m
2
(−iZ∇)
Eψ(r) = −4iZm
2
∇ψ(r) + V(r)ψ(r)
9.20 (a)
(b) Δ
Δ
x
m
.
( .
=
=
×
×
−
−
Z
9
2
1 05 10
2 1 67 10
34
27
Js
kg) )(
)
10
314
10
1
−
−
=
m s
m
Δ
Δ
x
m
.
( .
=
=
×
×
−
−
Z
9
2
1 05 10
2 1 67 10
34
27
Js
kg) )(
)
.
10
31 4
9
1
−
−
=
m s
m
ψ ψ τ
* d
V o
∫
λ
( .
)( .
)
=
=
×
×
−
−
hc
Φ
6 62 10
2 99 10
2
34
8
1
Js
m s
. .
( .
)
.
5
16 10
1
4 95 10
19
7
eV
J/ eV
m
×
×
=
×
−
−
λ
( .
)( .
)
.
=
=
×
×
−
−
hc
Φ
6 62 10
2 99 10
2 0
34
8
1
Js
m s
e eV
J/ eV
m
×
×
=
×
−
−
( .
)
.
1 6 10
1
6 21 10
19
7
λ
( .
)( .
)
=
=
×
×
−
−
hc
Φ
6 62 10
2 99 10
2
34
8
1
Js
m s
. .
(.
)
.
25
1 6 10
1
5 52 10
19
7
eV
J/ eV
m
×
×
=
×
−
−
KE
hc
( .
)( .
=
− =
×
×
−
λ
Φ
6 62 10
2 99 10
34
8
Js
m s s
m
eV
J
−
−
−
×
×
×
1
7
1 9
3 5 10
1
1 60 10
)
.
.
KE
hc
( .
)( .
=
− =
×
×
−
λ
Φ
6 62 10
2 99 10
34
8
Js
m s s
m
eV
J
−
−
−
×
×
×
1
7
1 9
2 5 10
1
1 60 10
)
.
.
KE
hc
( .
)( .
=
− =
×
×
−
λ
Φ
6 62 10
2 99 10
34
8
Js
m s s
m
eV
J
−
−
−
×
×
×
1
7
1 9
2 10
1
1 60 10
)
.
KE
hc
( .
)( .
=
− =
×
×
−
λ
Φ
6 62 10
2 99 10
34
8
Js
m s s
m
eV
J
−
−
−
×
×
×
1
7
1 9
3 5 10
1
1 60 10
)
.
.
KE
hc
( .
)( .
=
− =
×
×
−
λ
Φ
6 62 10
2 99 10
34
8
Js
m s s
m
eV
J
−
−
−
×
×
×
1
7
1 9
2 5 10
1
1 60 10
)
.
.
456
ANSWERS TO PROBLEMS
9781405124362_5_end.qxd 4/29/08 9:17 Page 456