(b)−2.0 eV =2.97 eV =4.75 × 10 −^19 J(c)−2.0 eV =1.55 eV =2.47 × 10 −^19 J9.14 (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 J9.15 (a)(b)(c)9.16The minimal frequency corresponds to a minimal energy (hν) needed
to overcome the work function.9.179.18According 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= 4 m^2 p→ 4 m^2 (−iZ∇)
Eψ(r) =− 4 iZm^2 ∇ψ(r) +V(r)ψ(r)9.20 (a)(b)
Δ
Δx
m.
(.
==
×
×
−
−Z
2 9
105 10
2 1 67 10
34
27Js
kg))( ) 10−− 10 1 =^314
msmΔ
Δ
x
m.
(.
==
×
×
−
−Z
2 9
105 10
2 1 67 10
34
27Js
kg))( ).
10
−− 91 =31 4
msmψψ τ*d
Vo∫
λ(. )(. )
==
hc ××−−
Φ6 62 10 2 99 10
2
(^34) Js (^8) m s 1
..(. )
.
516101
19 495 10^7
eV J/ eVm
××− =×−
λ(. )(. )
.
==
hc ××−−
Φ6 62 10 2 99 10
20
(^34) Js (^8) m s 1
eeV J/ eV
m
××
− =×−
(. )
.
16 10 1
19 621 10^7
λ(. )(. )
==
hc ××−−
Φ6 62 10 2 99 10
2
(^34) Js (^8) m s 1
..(. )
.
25 1 6 10 1
19 552 10^7
eV J/ eVm
××− =×−
KE
hc (. )(.
=−=××−
λΦ
6 62 10^34 Js 2 99 10^8 mss
meV
J−
× −−× ×1
35 10^7191
160 10
)
..
KE
hc (. )(.
=−=××−
λΦ
6 62 10^34 Js 2 99 10^8 mss
meV
J−
× −−× ×1
25 10^7191
160 10
)
..
KE
hc (. )(.
=−=××−
λΦ
6 62 10^34 Js 2 99 10^8 mss
meV
J−
× −−× ×1
2107191
160 10
)
.
KE
hc (. )(.
=−=××−
λΦ
6 62 10^34 Js 2 99 10^8 mss
meV
J−
× −−× ×1
35 10^7191
160 10
)
..
KE
hc (. )(.
=−=××−
λΦ
6 62 10^34 Js 2 99 10^8 mss
meV
J−
× −−× ×1
25 10^7191
160 10
)
..
456 ANSWERS TO PROBLEMS
9781405124362_5_end.qxd 4/29/08 9:17 Page 456