Chapter 27
Quantum Physics
Answers to Even Numbered Conceptual Questions
4.
Measuring the position of a particle implies having photons reflect from it. However,
collisions between photons and the particle will alter the velocity of the particle.
6.
Light has both wave and particle characteristics. In Young’s double-slit experiment, light
behaves as a wave. In the photoelectric effect, it behaves like a particle. Light can be
characterized as an electromagnetic wave with a particular wavelength or frequency,
yet at the same time, light can be characterized as a stream of photons, each carrying a
10.
Ultraviolet light has a shorter wavelength and higher photon energy than visible light.
16.
The red beam. Each photon of red light has less energy (longer wavelength) than a photon
of blue light, so the red beam must contain more photons to carry the same total energy.
393
394
CHAPTER 27
Problem Solutions
27.3
The wavelength of maximum radiation is given by
λm ax =
27.4
0.2898× 10− 2 m ⋅ K
= 5.00× 10−7 m = 500 nm
5800 K
The energy of a photon having wavelength λ is Eγ = hf = hc λ . Thus, the number of
photons delivered by each beam must be:
Red Beam:
( 2 500 eV ) ( 690× 10−9 m ) ⎛ 1.60× 10−19 J⎞
Etotal EtotalλR
3
=
=
nR =
⎜
⎟ = 1.39× 10
−34
8
1
eV
Eγ ,R
hc
6.
63
×
10
J
⋅
s
3.
00
×
10
m
s
(
)(
)⎝
⎠
( 2 500 eV ) ( 420× 10−9 m ) ⎛ 1.60× 10−19 J⎞
Etotal EtotalλB
=
=
= 845
nB =
Eγ ,B
hc
( 6.63× 10−34 J⋅ s)( 3.00× 108 m s) ⎜⎝ 1 eV ⎟⎠
Blue Beam:
27.8
The energy entering the eye each second is
P = I⋅ A = ( 4.0× 10−11 W m
2
) ⎡⎢⎣ π4 ( 8.5× 10
−3
m
)
2
⎤
−15
⎥⎦ = 2.3× 10 W
The energy of a single photon is
Eγ =
hc
λ
( 6.63× 10
=
− 34
J⋅ s) ( 3.00 × 108 m s)
−9
500 × 10 m
= 3.98× 10−19 J
so the number of photons entering the eye in ∆t= 1.00 s is
−15
∆E P⋅ ( ∆t) ( 2.3× 10 J s) ( 1.00 s)
N =
=
=
= 5.7 × 103
Eγ
Eγ
3.98 × 10−19 J
Quantum Physics
27.11
395
(a) From the photoelectric effect equation, the work function is
φ=
hc
λ
− KEm ax , or
( 6.63× 10
φ=
J⋅ s) ( 3.00 × 108 m s) ⎛
⎞
1 eV
⎜
⎟ − 1.31 eV
− 19
−9
350× 10 m
⎝ 1.60× 10 J⎠
− 34
φ = 2.24 eV
(b) λc =
(c)
27.19
fc =
hc
φ
c
λc
( 6.63× 10
− 34
=
=
J⋅ s) ( 3.00 × 108 m s) ⎛
⎞
1 eV
⎜
⎟ = 555 nm
− 19
2.24 eV
⎝ 1.60 × 10 J⎠
3.00 × 108 m s
= 5.41× 1014 H z
−9
555× 10 m
Assuming the electron produces a single photon as it comes to rest, the energy of that
photon is Eγ = ( KE )i = eV . The accelerating voltage is then
−34
8
hc ( 6.63× 10 J⋅ s)( 3.00 × 10 m s) 1.24× 10−6 V ⋅ m
V =
=
=
=
e eλ
λ
(1.60× 10-19 C ) λ
Eγ
For λ = 1.0 × 10−8 m , V =
1.24× 10−6 V ⋅ m
= 1.2× 102 V
1.0× 10−8 m
and for λ = 1.0× 10−13 m , V =
27.25
1.24× 10−6 V ⋅ m
= 1.2× 107 V
1.0 × 10−13 m
The interplanar spacing in the crystal is given by Bragg’s law as
d=
( 1)( 0.140 nm )
mλ
=
= 0.281 nm
2sinθ
2sin14.4°
396
CHAPTER 27
27.26
The scattering angle is given by the Compton shift formula as
⎛
θ = cos−1 ⎜ 1−
⎝
λC =
∆λ ⎞
⎟ where the Compton wavelength is
λC ⎠
h
= 0.002 43 nm
m ec2
⎛
1.50 × 10−3 nm ⎞
Thus, θ = cos−1 ⎜ 1−
⎟ = 67.5°
2.43× 10−3 nm ⎠
⎝
27.34
The de Broglie wavelength of a particle of mass m is λ = h p where the momentum is
given by p = γ m v = m v
1 − ( v c) . Note that when the particle is not relativistic, then
2
γ ≈ 1, and this relativistic expression for momentum reverts back to the classical
expression.
(a) For a proton moving at speed v = 2.00 × 104 m s , v << c and γ ≈ 1 so
λ=
h
6.63× 10−34 J⋅ s
=
= 1.98× 10−11 m
-27
4
m pv ( 1.67 × 10 kg )( 2.00× 10 m s)
(b) For a proton moving at speed v = 2.00× 107 m s
λ=
h
γ m pv
=
h
2
1− ( v c)
m pv
( 6.63× 10 J⋅ s)
=
(1.67 × 10 kg)( 2.00× 10
−34
-27
27.36
7
2
m s)
⎛ 2.00 × 107 m s⎞
−14
1− ⎜
⎟ = 1.98 × 10 m
8
3.
00
×
10
m
s
⎝
⎠
After falling freely with acceleration ay = − g = −9.80 m s2 for 50.0 m, starting from rest,
the speed of the ball will be
v = v02 + 2ay ( ∆y) = 0 + 2( −9.80 m s) ( −50.0 m ) = 31.3 m s
2
so the de Broglie wavelength is
λ=
h h
6.63× 10−34 J⋅ s
=
=
= 1.06× 10−34 m
p m v ( 0.200 kg )( 31.3 m s)
Quantum Physics
27.43
From the uncertainty principle, the minimum uncertainty in the momentum of the
electron is
∆px =
h
6.63× 10−34 J⋅ s
=
= 5.3× 10−25 kg ⋅ m s
4π ( ∆x) 4π ( 0.10× 10−9 m )
so the uncertainty in the speed of the electron is
∆vx =
∆px 5.3× 10−25 kg ⋅ m s
=
= 5.8× 105 m s or ~ 106 m s
−31
m
9.11× 10 kg
Thus, if the speed is on the order of the uncertainty in the speed, then v ~ 106 m s
27.45
With ∆x = 5.00 × 10−7 m s, the minimum uncertainty in the speed is
∆vx =
∆px
h
6.63× 10−34 J⋅ s
≥
=
= 116 m s
m e 4π m e ( ∆x) 4π ( 9.11× 10-31 kg )( 5.00× 10-7 m )
397