M39 YOUN7066 13 ISM c39 tủ tài liệu training

39

PARTICLES BEHAVING AS WAVES

39.1.

IDENTIFY and SET UP: λ =

h
h
=
. For an electron, m = 9.11 × 10−31 kg. For a proton,
p mv

m = 1.67 × 10−27 kg.
EXECUTE: (a) λ =

6.63 × 10−34 J ⋅ s
(9.11 × 10

−31

6

kg)(4.70 × 10 m/s)

= 1.55 × 10−10 m = 0.155 nm

⎛m ⎞
⎛ 9.11 × 10−31 kg ⎞
1

= 8.46 × 10−14 m.
, so λp = λe ⎜ e ⎟ = (1.55 × 10−10 m) ⎜
⎜ 1.67 × 10−27 kg ⎟⎟
⎜ mp ⎟
m
⎝
⎠
⎝
⎠
EVALUATE: For the same speed the proton has a smaller de Broglie wavelength.
h
p2
hc
IDENTIFY and SET UP: For a photon, E = . For an electron or proton, p = and E =
, so
λ
λ
2m
(b) λ is proportional to

39.2.

E=

h2
2mλ 2

.

EXECUTE: (a) E =

hc

λ

=

(4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s)

= 6.2 keV

2

⎛ 6.63 × 10−34 J ⋅ s ⎞
1
=
= 6.03 × 10−18 J = 38 eV
(b) E =
⎜⎜
⎟⎟
2
−9
−31
2mλ
⎝ 0.20 × 10 m ⎠ 2(9.11 × 10 kg)
⎛m ⎞
⎛ 9.11 × 10−31 kg ⎞
= 0.021 eV
(c) Ep = Ee ⎜ e ⎟ = (38 eV) ⎜
⎜ 1.67 × 10−27 kg ⎟⎟

⎜ mp ⎟
⎝
⎠
⎝
⎠
EVALUATE: For a given wavelength a photon has much more energy than an electron, which in turn has
more energy than a proton.
h
p2
IDENTIFY: For a particle with mass, λ =
and K =
.
2m
p
h2

39.3.

0.20 × 10−9 m

SET UP: 1 eV = 1.60 × 10−19 J
EXECUTE: (a) λ =

h
h (6.63 × 10−34 J ⋅ s)
⇒ p= =
= 2.37 × 10−24 kg ⋅ m/s.
p
λ (2.80 × 10−10 m)

p 2 (2.37 × 10−24 kg ⋅ m/s) 2
=
= 3.08 × 10−18 J = 19.3 eV.
2m
2(9.11 × 10−31 kg)
EVALUATE: This wavelength is on the order of the size of an atom. This energy is on the order of the
energy of an electron in an atom.
(b) K =

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39-1


39-2

39.4.

Chapter 39

IDENTIFY: For a particle with mass, λ =

p2
h
and E =
.
2m
p

SET UP: 1 eV = 1.60 × 10−19 J
EXECUTE: λ =

39.5.

h
h
(6.63 × 10−34 J ⋅ s)
=
=
= 7.02 × 10−15 m.
−27
6
−19
p
2mE
2(6.64 × 10 kg) (4.20 × 10 eV) (1.60 × 10 J/eV)

EVALUATE: This wavelength is on the order of the size of a nucleus.
h
h
IDENTIFY and SET UP: The de Broglie wavelength is λ = =
. In the Bohr model, mvrn = n(h /2π ),
p mv

so mv = nh /(2π rn ). Combine these two expressions and obtain an equation for λ in terms of n. Then
⎛ 2π rn ⎞ 2π rn
.
⎟=
n

⎝ nh ⎠

λ = h⎜

EXECUTE: (a) For n = 1, λ = 2π r1 with r1 = a0 = 0.529 × 10 −10 m, so

λ = 2π (0.529 × 10−10 m) = 3.32 × 10−10 m.
λ = 2π r1; the de Broglie wavelength equals the circumference of the orbit.
(b) For n = 4, λ = 2π r4 /4.
rn = n 2 a0 so r4 = 16a0 .

λ = 2π (16a0 )/4 = 4(2π a0 ) = 4(3.32 × 10−10 m) = 1.33 × 10−9 m
1 1
λ = 2π r4 /4; the de Broglie wavelength is = times the circumference of the orbit.
n 4
EVALUATE: As n increases the momentum of the electron increases and its de Broglie wavelength
decreases. For any n, the circumference of the orbits equals an integer number of de Broglie wavelengths.
39.6.

IDENTIFY: λ =

h
p

SET UP: 1 eV = 1.60 × 10−19 J. An electron has mass 9.11 × 10−31 kg.
EXECUTE: (a) For a nonrelativistic particle, K =

p2
h
, so λ = =

2m
p

h
.
2 Km

(b) (6.63 × 10−34 J ⋅ s) / 2(800 eV)(1.60 × 10−19 J/eV)(9.11 × 10−31 kg) = 4.34 × 10−11 m.
39.7.

EVALUATE: The de Broglie wavelength decreases when the kinetic energy of the particle increases.
IDENTIFY: A person walking through a door is like a particle going through a slit and hence should
exhibit wave properties.
SET UP: The de Broglie wavelength of the person is λ = h /mv.
EXECUTE: (a) Assume m = 75 kg and v = 1.0 m/s.

λ = h /mv = (6.626 × 10−34 J ⋅ s)/[(75 kg)(1.0 m/s)] = 8.8 × 10−36 m
EVALUATE: (b) A typical doorway is about 1 m wide, so the person’s de Broglie wavelength is much too
small to show wave behavior through a “slit” that is about 1035 times as wide as the wavelength. Hence
ordinary objects do not show wave behavior in everyday life.
39.8.

IDENTIFY and SET UP: Combining Eqs. 37.38 and 37.39 gives p = mc γ 2 − 1.
EXECUTE: (a) λ =

h
= ( h /mc)/ γ 2 − 1 = 4.43 × 10−12 m. (The incorrect nonrelativistic calculation gives
p

5.05 × 10−12 m. )

(b) (h /mc)/ γ 2 − 1 = 7.07 × 10−13 m.
EVALUATE: The de Broglie wavelength decreases when the speed increases.

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Particles Behaving as Waves
39.9.

39-3

IDENTIFY and SET UP: A photon has zero mass and its energy and wavelength are related by Eq. (38.2).
An electron has mass. Its energy is related to its momentum by E = p 2 /2m and its wavelength is related to

its momentum by Eq. (39.1).
hc
hc (6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s)
EXECUTE: (a) photon: E =
so λ =
=
= 62.0 nm.
λ
E
(20.0 eV)(1.602 × 10−19 J/eV)

electron: E = p 2 /(2m) so p = 2mE =

2(9.109 × 10−31 kg)(20.0 eV)(1.602 × 10−19 J/eV) = 2.416 × 10−24 kg ⋅ m/s. λ = h /p = 0.274 nm.
(b) photon: E = hc/R = 7.946 × 10−19 J = 4.96 eV.

electron: λ = h /p so p = h /λ = 2.650 × 10−27 kg ⋅ m/s.
E = p 2 /(2m) = 3.856 × 10−24 J = 2.41 × 10−5 eV.

(c) EVALUATE: You should use a probe of wavelength approximately 250 nm. An electron with
λ = 250 nm has much less energy than a photon with λ = 250 nm, so is less likely to damage the
molecule. Note that λ = h /p applies to all particles, those with mass and those with zero mass.
E = hf = hc /λ applies only to photons and E = p 2 /2m applies only to particles with mass.

39.10.

IDENTIFY: Knowing the de Broglie wavelength for an electron, we want to find its speed.
h
h
= 1.00 mm, m = 9.11 × 10−31 kg.
SET UP: λ = =
p mv

h
6.63 × 10−34 J ⋅ s
=
= 0.728 m/s.
mλ (9.11 × 10−31 kg)(1.00 × 10−3 m)
EVALUATE: Electrons normally move much faster than this, so their de Broglie wavelengths are much
much smaller than a millimeter.
IDENTIFY and SET UP: Use Eq. (39.1).
h
h
6.626 × 10−34 J ⋅ s
EXECUTE: λ = =

=
= 3.90 × 10−34 m
p mv (5.00 × 10−3 kg)(340 m/s)
EXECUTE: v =

39.11.

39.12.

EVALUATE: This wavelength is extremely short; the bullet will not exhibit wavelike properties.
IDENTIFY: The kinetic energy of the electron is equal to the energy of the photon. We want to find the
wavelengths of each of them.
SET UP: Both for particles with mass (electrons) and for massless particles (photons) the wavelength is
h
related to the momentum p by λ = . But for each type of particle there is a different expression that
p

relates the energy E and momentum p. For an electron E = 12 mv 2 =
EXECUTE: photon: p =

p2
but for a photon E = pc.
2m

E
h
h E
hc 1.24 × 10−6 eV ⋅ m
and p = so =
and λ =

=
= 49.6 nm.
λ
λ c
c
E
25 eV

electron: Solving for p gives p = 2mE . This gives

p = 2(9.11 × 10−31 kg)(25 eV)(1.60 × 10−19 J/eV) = 2.70 × 10−24 kg ⋅ m/s. The wavelength is therefore

λ=

h
6.63 × 10−34 J ⋅ s
=
= 0.245 nm.
p 2.70 × 10−24 kg ⋅ m/s

EVALUATE: The wavelengths are quite different. For the electron λ =

h
hc
and for the photon λ = ,
E
2mE

so for an electron λ is proportional to E −1/2 and for a photon λ is proportional to E −1. It is incorrect to
say p =

E
for a particle such as an electron that has mass; the correct relation is p =
c

E 2 − ( mc 2 ) 2
.
c

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39-4

Chapter 39

39.13.

IDENTIFY: The acceleration gives momentum to the electrons. We can use this momentum to calculate
their de Broglie wavelength.
SET UP: The kinetic energy K of the electron is related to the accelerating voltage V by K = eV . For an

electron E = 12 mv 2 =

h
p2
hc
and λ = . For a photon E = .
λ

2m
p

EXECUTE: (a) For an electron p =

h

λ

=

6.63 × 10−34 J ⋅ s
5.00 × 10−9 m

= 1.33 × 10−25 kg ⋅ m/s and

p 2 (1.33 × 10−25 kg ⋅ m/s) 2
K 9.71 × 10−21 J
=
= 9.71 × 10−21 J. V = =
= 0.0607 V. The electrons
−
31
2m
e 1.60 × 10−19 C
2(9.11 × 10 kg)
would have kinetic energy 0.0607 eV.
E=

(b) E =

hc

λ

=

1.24 × 10−6 eV ⋅ m
5.00 × 10−9 m

= 248 eV.

(c) E = 9.71 × 10−21 J

so λ =

39.14.

hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s)
=
= 20.5 μ m.
E
9.71 × 10−21 J

EVALUATE: If they have the same wavelength, the photon has vastly more energy than the electron.
h
IDENTIFY: λ = . Apply conservation of energy to relate the potential difference to the speed of the
p
electrons.
hc

SET UP: The mass of an electron is m = 9.11 × 10−31 kg. The kinetic energy of a photon is E = .

λ

1
EXECUTE: (a) λ = h /mv → v = h /mλ . Energy conservation: eΔV = mv 2 .
2
2

⎛ h ⎞
m⎜
⎟
mv 2
h2
(6.626 × 10−34 J ⋅ s) 2
mλ ⎠
ΔV =
= ⎝
=
=
= 66.9 V.
−
2
19
2e
2e
2emλ
2(1.60 × 10 C)(9.11 × 10−31 kg) (0.15 × 10−9 m) 2
(b) Ephoton = hf =

ΔV =

Ephoton
e

=

hc

λ

=

(6.626 × 10−34 J ⋅ s) (3.0 × 108 m/s)

1.33 × 10−15 J
1.6 × 10−19 C

0.15 × 10−9 m

= 8310 V.

EVALUATE: The electron in part (b) has wavelength λ =

39.15.

= 1.33 × 10−15 J. eΔV = K = Ephoton and

h
=

p

h
= 0.0135 nm, much shorter than the
2mE

wavelength of a photon of the same energy.
h
hc
IDENTIFY: For an electron, λ = and K = 12 mv 2 . For a photon, E = . The wavelength should be 0.10 nm.
p
λ
SET UP: For an electron, m = 9.11 × 10−31 kg.
EXECUTE: (a) λ = 0.10 nm. p = mv = h /λ so v = h /(mλ ) = 7.3 × 106 m/s.
(b) K =

1 2
mv = 150 eV.
2

(c) E = hc /λ = 12 keV.
EVALUATE: (d) The electron is a better probe because for the same λ it has less energy and is less
damaging to the structure being probed.

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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


Particles Behaving as Waves
39.16.

39-5

IDENTIFY: The electrons behave like waves and are diffracted by the slit.
SET UP: We use conservation of energy to find the speed of the electrons, and then use this speed to find
their de Broglie wavelength, which is λ = h /mv. Finally we know that the first dark fringe for single-slit
diffraction occurs when a sin θ = λ.
EXECUTE: (a) Use energy conservation to find the speed of the electron:

1
2

mv 2 = eV .

2eV
2(1.60 × 10−19 C)(100 V)
=
= 5.93 × 106 m/s
m
9.11 × 10−31 kg

v=

which is about 2% the speed of light, so we can ignore relativity.
(b) First find the de Broglie wavelength:

λ=

h
6.626 × 10−34 J ⋅ s

=
= 1.23 × 10−10 m = 0.123 nm
mv (9.11 × 10−31 kg)(5.93 × 106 m/s)

For the first single-slit dark fringe, we have a sin θ = λ , which gives
a=

39.17.

λ
sin θ

=

1.23 × 10−10 m
= 6.16 × 10−10 m = 0.616 nm
sin(11.5°)

EVALUATE: The slit width is around 5 times the de Broglie wavelength of the electron, and both are much
smaller than the wavelength of visible light.
h
IDENTIFY: The intensity maxima are located by Eq. (39.4). Use λ =
for the wavelength of the
p

neutrons. For a particle, p = 2mE .
SET UP: For a neutron, m = 1.67 × 10−27 kg.
EXECUTE: For m = 1, λ = d sin θ =

E=

h2
2md 2 sin 2 θ

=

h
.
2mE

(6.63 × 10−34 J ⋅ s)2
2(1.675 × 10−27 kg) (9.10 × 10−11 m) 2 sin 2 (28.6°)

= 6.91 × 10−20 J = 0.432 eV.

EVALUATE: The neutrons have λ = 0.0436 nm, comparable to the atomic spacing.
39.18.

IDENTIFY: Intensity maxima occur when d sin θ = mλ . λ =

h
h
mh
=
so d sin θ =
.
p
2ME
2ME

SET UP: Here m is the order of the maxima, whereas M is the mass of the incoming particle.
EXECUTE: (a) d =

(2)(6.63 × 10−34 J ⋅ s)
mh
=
=
2ME sin θ
2(9.11 × 10−31 kg)(188 eV)(1.60 × 10−19 J/eV) sin(60.6°)

2.06 × 10−10 m = 0.206 nm.
(b) m = 1 also gives a maximum.
⎛
⎞
(1)(6.63 × 10−34 J ⋅ s)
⎟ = 25.8°. This is the only other
θ = arcsin ⎜
⎜ 2(9.11 × 10−31 kg)(188 eV)(1.60 × 10−19 J/eV)(2.06 × 10−10 m) ⎟
⎝
⎠
one. If we let m ≥ 3, then there are no more maxima.
(c) E =

m2h2
2 Md 2 sin 2 θ

=

(1) 2 (6.63 × 10−34 J ⋅ s) 2
2(9.11 × 10−31 kg) (2.60 × 10−10 m) 2 sin 2 (60.6°)

.

= 7.49 × 10−18 J = 46.8 eV.
Using this energy, if we let m = 2, then sin θ > 1. Thus, there is no m = 2 maximum in this case.
EVALUATE: As the energy of the electrons is lowered their wavelength increases and a given intensity
maximum occurs at a larger angle.
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39-6

Chapter 39

39.19.

IDENTIFY: The condition for a maximum is d sin θ = mλ . λ =

h
h
⎛ mh ⎞
=
, so θ = arcsin ⎜
⎟.
p Mv
⎝ dMv ⎠

SET UP: Here m is the order of the maximum, whereas M is the incoming particle mass.
⎛ h ⎞

EXECUTE: (a) m = 1 ⇒ θ1 = arcsin ⎜
⎟
⎝ dMv ⎠

⎛
⎞
6.63 × 10−34 J ⋅ s
= arcsin ⎜
= 2.07°.
⎜ (1.60 × 10−6 m)(9.11 × 10−31 kg)(1.26 × 104 m/s) ⎟⎟
⎝
⎠
⎛
⎞
(2)(6.63 × 10−34 J ⋅ s)
= 4.14°.
m = 2 ⇒ θ 2 = arcsin ⎜
⎜ (1.60 × 10−6 m)(9.11 × 10−31 kg)(1.26 × 104 m/s) ⎟⎟
⎝
⎠
⎛ π radians ⎞
(b) For small angles (in radians!) y ≅ Dθ , so y1 ≈ (50.0 cm) (2.07°) ⎜
⎟ = 1.81 cm,
⎝ 180° ⎠
⎛ π radians ⎞
y2 ≈ (50.0 cm) (4.14°) ⎜
⎟ = 3.61 cm and y2 − y1 = 3.61 cm − 1.81 cm = 1.80 cm.
⎝ 180° ⎠
EVALUATE: For these electrons, λ =

h
= 0.0577 μm. λ is much less than d and the intensity maxima
mv

occur at small angles.
39.20.

IDENTIFY: λ =

h
p2
. Conservation of energy gives eV = K =
, where V is the accelerating voltage.
p
2m

SET UP: The electron mass is 9.11 × 10−31 kg and the proton mass is 1.67 × 10−27 kg.
EXECUTE: (a) eV = K =

p 2 ( h /λ ) 2
(h /λ ) 2
=
= 419 V.
, so V =
2m
2m
2me

(b) The voltage is reduced by the ratio of the particle masses, (419 V)

9.11 × 10−31 kg
1.67 × 10−27 kg

= 0.229 V.

h
h
=
. For the same λ , particles of greater mass have smaller E, so a smaller
p
2mE
accelerating voltage is needed for protons.
EVALUATE: λ =

39.21.

IDENTIFY and SET UP: For a photon Eph =

hc

λ

=

1.99 × 10−25 J ⋅ m

λ

. For an electron

2

Ee =

p2
1 ⎛h⎞
h2
.
=
=
⎜ ⎟
2m 2m ⎝ λ ⎠
2mλ 2

EXECUTE: (a) photon Eph =

electron Ee =

Eph

1.99 × 10−25 J ⋅ m
10.0 × 10−9 m

= 1.99 × 10−17 J

(6.63 × 10−34 J ⋅ s) 2
2(9.11 × 10−31 kg)(10.0 × 10−9 m) 2

= 2.41 × 10−21 J

1.99 × 10−17 J

= 8.26 × 103
2.41 × 10−21 J
(b) The electron has much less energy so would be less damaging.

Ee

=

EVALUATE: For a particle with mass, such as an electron, E ~ λ −2 . For a massless photon E ~ λ −1.

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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


Particles Behaving as Waves
39.22.

39-7

IDENTIFY: The kinetic energy of the alpha particle is all converted to electrical potential energy at closest
approach. The force on the alpha particle is the electrical repulsion of the nucleus.
1 q1q2
SET UP: The electrical potential energy of the system is U =
and the kinetic energy is
4πε 0 r

K = 12 mv 2 . The electrical force is R = 2.5 m (at closest approach).
(a) Equating the initial kinetic energy and the final potential energy and solving for the separation radius

r gives

1 (92e) (2e)
1
(184) (1.60 × 10−19 C) 2
=
= 5.54 × 10−14 m.
4πε 0
4πε 0 (4.78 × 106 eV)(1.60 × 10−19 J/eV)
K
(b) The above result may be substituted into Coulomb’s law. Alternatively, the relation between the
U
magnitude of the force and the magnitude of the potential energy in a Coulomb field is F =
. U = K,
r

r=

so F =

39.23.

K (4.78 × 106 eV) (1.6 × 10−19 J/ev)
=
= 13.8 N.
r
(5.54 × 10−14 m)

EVALUATE: The result in part (a) is comparable to the radius of a large nucleus, so it is reasonable. The
force in part (b) is around 3 pounds, which is large enough to be easily felt by a person.

1 q1q2
(a) IDENTIFY: If the particles are treated as point charges, U =
.
4π ⑀0 r
SET UP: q1 = 2e (alpha particle); q2 = 82e (gold nucleus); r is given so we can solve for U.
EXECUTE: U = (8.987 × 109 N ⋅ m 2 /C 2 )

(2)(82)(1.602 × 10−19 C) 2
6.50 × 10−14 m

= 5.82 × 10−13 J

U = 5.82 × 10−13 J (1 eV/1.602 × 10−19 J) = 3.63 × 106 eV = 3.63 MeV
(b) IDENTIFY: Apply conservation of energy: K1 + U1 = K 2 + U 2 .
SET UP: Let point 1 be the initial position of the alpha particle and point 2 be where the alpha particle
momentarily comes to rest. Alpha particle is initially far from the lead nucleus implies r1 ≈ ∞ and U1 = 0.

Alpha particle stops implies K 2 = 0.
EXECUTE: Conservation of energy thus says K1 = U 2 = 5.82 × 10−13 J = 3.63 MeV.
(c) K =

39.24.

2K
2(5.82 × 10−13 J)
1 2
mv so v =
=
= 1.32 × 107 m/s
2

m
6.64 × 10−27 kg

EVALUATE: v /c = 0.044, so it is ok to use the nonrelativistic expression to relate K and v. When the alpha
particle stops, all its initial kinetic energy has been converted to electrostatic potential energy.
IDENTIFY: The minimum energy the photon would need is the 3.84 eV bond strength.
hc
SET UP: The photon energy E = hf =
must equal the bond strength.

λ

hc
(4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s)
=
= 327 nm.
λ
3.80 eV
3.80 eV
EVALUATE: Any photon having a shorter wavelength would also spell doom for the Horta!
IDENTIFY and SET UP: Use the energy to calculate n for this state. Then use the Bohr equation, Eq. (39.6),
to calculate L.
EXECUTE: En = −(13.6 eV)/n 2 , so this state has n = 13.6/1.51 = 3. In the Bohr model, L = n so for
EXECUTE:

39.25.

hc

= 3.80 eV, so λ =

this state L = 3ћ = 3.16 × 10−34 kg ⋅ m 2 /s.
EVALUATE: We will find in Section 41.1 that the modern quantum mechanical description gives a different result.

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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


39-8

39.26.

Chapter 39

13.6 eV

hc
. ΔE = , where ΔE is the magnitude of
λ
n2
the energy change for the atom and λ is the wavelength of the photon that is absorbed or emitted.

IDENTIFY and SET UP: For a hydrogen atom En = −

⎛ 1 1⎞
EXECUTE: ΔE = E4 − E1 = −(13.6 eV) ⎜ 2 − 2 ⎟ = +12.75 eV.
⎝4 1 ⎠

λ=

hc (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s)
c
=
= 97.3 nm. f = = 3.08 × 1015 Hz.
λ
ΔE
12.75 eV

EVALUATE: This photon is in the ultraviolet region of the electromagnetic spectrum.
39.27.

IDENTIFY: The force between the electron and the nucleus in Be3+ is F =

1

Ze2

4π ⑀0 r 2

, where Z = 4 is the

nuclear charge. All the equations for the hydrogen atom apply to Be3+ if we replace e 2 by Ze 2 .
(a) SET UP: Modify Eq. (39.14).
EXECUTE: En = −

En = −

1 m( Ze 2 ) 2

⑀02 8n 2h 2

1 me 4

⑀02 8n 2h 2

(hydrogen) becomes

⎛ 1 me 4 ⎞
⎛ 13.60 eV ⎞
3+
= Z2⎜− 2 2 2 ⎟ = Z2⎜−
⎟ (for Be )
2
⎜ ⑀ 8n h ⎟
n
⎝
⎠
⎝ 0
⎠

⎛ 13.60 eV ⎞
The ground-level energy of Be3+ is E1 = 16 ⎜ −
⎟ = −218 eV.
12
⎝
⎠
EVALUATE: The ground-level energy of Be3+ is Z 2 = 16 times the ground-level energy of H.
(b) SET UP: The ionization energy is the energy difference between the n → ∞ level energy and the
n = 1 level energy.
EXECUTE: The n → ∞ level energy is zero, so the ionization energy of Be3+ is 218 eV.

EVALUATE: This is 16 times the ionization energy of hydrogen.
(c) SET UP:

⎛ 1
1 ⎞
= R ⎜ 2 − 2 ⎟ just as for hydrogen but now R has a different value.
⎜
⎟
λ
⎝ n1 n2 ⎠
1

EXECUTE: RH =

RBe = Z 2

me4
8⑀02 h3c

me4
8⑀02 h3c

= 1.097 × 107 m −1 for hydrogen becomes

= 16(1.097 × 107 m −1 ) = 1.755 × 108 m −1 for Be3+ .

For n = 2 to n = 1,

⎛1 1 ⎞
= RBe ⎜ 2 − 2 ⎟ = 3RBe /4.

λ
⎝1 2 ⎠
1

λ = 4/(3RBe ) = 4/(3(1.755 × 108 m −1)) = 7.60 × 10−9 m = 7.60 nm.
EVALUATE: This wavelength is smaller by a factor of 16 compared to the wavelength for the
corresponding transition in the hydrogen atom.
(d) SET UP: Modify Eq. (39.8): rn = ⑀ 0
EXECUTE: rn = ⑀ 0

n2h2

π m( Ze2 )

n2h2

π me2

(hydrogen).

(Be3+ ).

EVALUATE: For a given n the orbit radius for Be3+ is smaller by a factor of Z = 4 compared to the
corresponding radius for hydrogen.

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Particles Behaving as Waves

39.28.

IDENTIFY and SET UP: En = −
EXECUTE: (a) En = −

13.6 eV

n2

13.6 eV

n2

39-9

.

and En +1 = −

13.6 eV
(n + 1) 2

.

⎡ 1
1⎤
n 2 − ( n + 1) 2
2n + 1
ΔE = En +1 − En = ( −13.6 eV) ⎢

− 2 ⎥ = − (13.6 eV) 2
. ΔE = (13.6 eV) 2
.
2
2
+
+
(
1)
(
)(
1)
n
n
n
n
(
n
)( n + 1) 2
⎣
⎦
2n
2
As n becomes large, ΔE → (13.6 eV) 4 = (13.6 eV) 3 . Thus ΔE becomes small as n becomes large.
n
n
(b) rn = n 2 r1 so the orbits get farther apart in space as n increases.
EVALUATE: There are an infinite number of bound levels for the hydrogen atom. As n increases the
energies of the bound levels converge to the ionization threshold.
39.29.

IDENTIFY: Apply Eqs. (39.8) and (39.9).
SET UP: The orbital period for state n is the circumference of the orbit divided by the orbital speed.
EXECUTE: (a) vn =
n = 2 ⇒ v2 =

v1
v
= 1.09 × 106 m/s. n = 3 ⇒ v3 = 1 = 7.27 × 105 m/s.
2
3

(b) Orbital period =

n = 1 ⇒ T1 =

1 e2
(1.60 × 10−19 C) 2
: n = 1 ⇒ v1 =
= 2.18 × 106 m/s.
ε 0 2nh
ε 0 2 (6.63 × 10−34 J ⋅ s)

2π rn 2ε 0 n 2 h 2 /me2 4ε 02 n3h3
.
=
=
vn
me4
1/ε 0 ⋅ e2 /2nh

4ε 02 (6.63 × 10−34 J ⋅ s)3

(9.11 × 10−31 kg)(1.60 × 10−19 C) 4

= 1.53 × 10−16 s

n = 2: T2 = T1 (2)3 = 1.22 × 10−15 s. n = 3: T3 = T1 (3)3 = 4.13 × 10−15 s.
(c) number of orbits =

1.0 × 10−8 s
1.22 × 10−15 s

= 8.2 × 106.

EVALUATE: The orbital speed is proportional to1/n, the orbital radius is proportional to n 2 , and the

orbital period is proportional to n3.
39.30.

IDENTIFY and SET UP: The ionization threshold is at E = 0. The energy of an absorbed photon equals the
energy gained by the atom and the energy of an emitted photon equals the energy lost by the atom.
EXECUTE: (a) ΔE = 0 − (−20 eV) = 20 eV
(b) When the atom in the n = 1 level absorbs an 18-eV photon, the final level of the atom is n = 4.
The possible transitions from n = 4 and corresponding photon energies are n = 4 → n = 3, 3 eV;
n = 4 → n = 2, 8 eV; n = 4 → n = 1, 18 eV. Once the atom has gone to the n = 3 level, the following
transitions can occur: n = 3 → n = 2, 5 eV; n = 3 → n = 1,15 eV. Once the atom has gone to the n = 2
level, the following transition can occur: n = 2 → n = 1, 10 eV. The possible energies of emitted photons

are: 3 eV, 5 eV, 8 eV, 10 eV, 15 eV and 18 eV.

(c) There is no energy level 8 eV higher in energy than the ground state, so the photon cannot be absorbed.
(d) The photon energies for n = 3 → n = 2 and for n = 3 → n = 1 are 5 eV and 15 eV. The photon energy
for n = 4 → n = 3 is 3 eV. The work function must have a value between 3 eV and 5 eV.
EVALUATE: The atom has discrete energy levels, so the energies of emitted or absorbed photons have
only certain discrete energies.

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39-10
39.31.

Chapter 39
IDENTIFY and SET UP: The wavelength of the photon is related to the transition energy Ei − Ef of the

atom by Ei − Ef =

hc

λ

where hc = 1.240 × 10−6 eV ⋅ m.

EXECUTE: (a) The minimum energy to ionize an atom is when the upper state in the transition has E = 0,

so E1 = −17.50 eV. For n = 5 → n = 1, λ = 73.86 nm and E5 − E1 =

1.240 × 10−6 eV ⋅ m
73.86 × 10−9 m

= 16.79 eV.

E5 = −17.50 eV + 16.79 eV = −0.71 eV. For n = 4 → n = 1, λ = 75.63 nm and E4 = −1.10 eV. For

n = 3 → n = 1, λ = 79.76 nm and E3 = −1.95 eV. For n = 2 → n = 1, λ = 94.54 nm and E2 = −4.38 eV.
(b) Ei − Ef = E4 − E2 = −1.10 eV − (−4.38 eV) = 3.28 eV and λ =

39.32.

hc
1.240 × 10−6 eV ⋅ m
=
= 378 nm
Ei − Ef
3.28 eV

EVALUATE: The n = 4 → n = 2 transition energy is smaller than the n = 4 → n = 1 transition energy so
the wavelength is longer. In fact, this wavelength is longer than for any transition that ends in the n = 1
state.
IDENTIFY and SET UP: For the Lyman series the final state is n = 1 and the wavelengths are given by
1
⎛1 1 ⎞
= R ⎜ 2 − 2 ⎟ , n = 2, 3,…. For the Paschen series the final state is n = 3 and the wavelengths are given
λ
⎝1 n ⎠

1 ⎞
⎛ 1
= R ⎜ 2 − 2 ⎟ , n = 4, 5,…. R = 1.097 × 107 m −1. The longest wavelength is for the smallest n and

λ
n ⎠
⎝3
the shortest wavelength is for n → ∞.
4
1
⎛ 1 1 ⎞ 3R
EXECUTE: Lyman: Longest: = R ⎜ 2 − 2 ⎟ =
= 121.5 nm.
. λ=
λ
3(1.097 × 107 m −1)
⎝1 2 ⎠ 4
by

1

Shortest:

1

λ

1
1 ⎞
⎛1
= R ⎜ 2 − 2 ⎟ = R. λ =
= 91.16 nm
1.097 × 107 m −1
⎝1 ∞ ⎠

Paschen: Longest:

1 ⎞ R
⎛ 1
= R⎜ 2 − 2 ⎟ = .
∞ ⎠ 9
⎝3
EVALUATE: The Lyman series is in the ultraviolet. The Paschen series is in the infrared.
IDENTIFY: Apply conservation of energy to the system of atom and photon.
hc
SET UP: The energy of a photon is Eγ = .

Shortest:
39.33.

144
1 ⎞ 7R
⎛ 1
= R⎜ 2 − 2 ⎟ =
= 1875 nm.
.λ=
λ
7(1.097 × 107 m −1 )
4 ⎠ 144
⎝3
1

1

λ

λ
hc (6.63 × 10
J ⋅ s)(3.00 × 108 m/s)
EXECUTE: (a) Eγ =
=
= 2.31 × 10−19 J = 1.44 eV. So the internal
λ
8.60 × 10−7 m
energy of the atom increases by 1.44 eV to E = −6.52 eV + 1.44 eV = −5.08 eV.
−34

(b) Eγ =

hc

λ

=

(6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s)
−7

= 4.74 × 10−19 J = 2.96 eV. So the final internal energy of

4.20 × 10 m
the atom decreases to E = −2.68 eV − 2.96 eV = −5.64 eV.

39.34.

EVALUATE: When an atom absorbs a photon the energy of the atom increases. When an atom emits a
photon the energy of the atom decreases.
1
1 ⎞
⎛ 1
IDENTIFY and SET UP: Balmer’s formula is = R ⎜ 2 − 2 ⎟ . For the Hγ spectral line n = 5. Once we
λ
n ⎠
⎝2
have λ , calculate f from f = c /λ and E from Eq. (38.2).
EXECUTE: (a)

1 ⎞
⎛ 1
⎛ 25 − 4 ⎞
⎛ 21 ⎞
= R⎜ 2 − 2 ⎟ = R⎜
⎟ = R⎜
⎟.
λ
5 ⎠
⎝2
⎝ 100 ⎠
⎝ 100 ⎠
1

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Particles Behaving as Waves

Thus λ =
(b) f =

c

λ

39-11

100
100
m = 4.341 × 10−7 m = 434.1 nm.
=
21R 21(1.097 × 107 )
=

2.998 × 108 m/s
4.341 × 10−7 m

= 6.906 × 1014 Hz

(c) E = hf = (6.626 × 10−34 J ⋅ s)(6.906 × 1014 Hz) = 4.576 × 10−19 J = 2.856 eV
EVALUATE: Section 39.3 shows that the longest wavelength in the Balmer series (Hα ) is 656 nm and the

shortest is 365 nm. Our result for Hγ falls within this range. The photon energies for hydrogen atom
39.35.

transitions are in the eV range, and our result is of this order.
IDENTIFY: We know the power of the laser beam, so we know the energy per second that it delivers. The
wavelength of the light tells us the energy of each photon, so we can use that to calculate the number of
photons delivered per second.
hc 1.99 × 10−25 J ⋅ m
. The power is the total energy per
SET UP: The energy of each photon is E = hf =
=

λ

λ

second and the total energy Etot is the number of photons N times the energy E of each photon.
EXECUTE: λ = 10.6 × 10−6 m, so E = 1.88 × 10−20 J. P =

39.36.

Etot NE
so
=
t
t

N P 0.100 × 103 W
= =
= 5.32 × 1021 photons/s.
t
E 1.88 × 10−20 J
EVALUATE: At over 1021 photons per second, we can see why we do not detect individual photons.

IDENTIFY: We can calculate the energy of a photon from its wavelength. Knowing the intensity of the
beam and the energy of a single photon, we can determine how many photons strike the blemish with each
pulse.
hc 1.99 × 10−25 J ⋅ m
SET UP: The energy of each photon is E = hf =
. The power is the total energy per
=

λ

λ

second and the total energy Etot is the number of photons N times the energy E of each photon. The
photon beam is spread over an area A = π r 2 with r = 2.5 mm.
EXECUTE: (a) λ = 585 nm and E =
(b) P =

hc

λ

= 3.40 × 10−19 J = 2.12 eV.

Etot NE
Pt (20.0 W)(0.45 × 10−3 s)
=
so N =
=
= 2.65 × 1016 photons. These photons are spread
t

t
E
3.40 × 10−19 J

over an area π r 2, so the number of photons per mm 2 is

39.37.

2.65 × 1016 photons

π (2.5 mm) 2

= 1.35 × 1015 photons/mm 2 .

EVALUATE: With so many photons per mm 2 , it is impossible to detect individual photons.
IDENTIFY and SET UP: The number of photons emitted each second is the total energy emitted divided by
the energy of one photon. The energy of one photon is given by Eq. (38.2). E = Pt gives the energy
emitted by the laser in time t.
EXECUTE: In 1.00 s the energy emitted by the laser is (7.50 × 10−3 W)(1.00 s) = 7.50 × 10−3 J.

The energy of each photon is E =

hc

λ

=

(6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s)
10.6 × 10−6 m

= 1.874 × 10−20 J.

7.50 × 10−3 J/s

= 4.00 × 1017 photons/s
1.874 × 10−20 J/photon
EVALUATE: The number of photons emitted per second is extremely large.

Therefore

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39-12
39.38.

Chapter 39
IDENTIFY and SET UP: Visible light has wavelengths from about 400 nm to about 700 nm. The energy of
hc 1.99 × 10−25 J ⋅ m
. The power is the total energy per second and the total
each photon is E = hf =
=

λ

λ

energy Etot is the number of photons N times the energy E of each photon.

EXECUTE: (a) 193 nm is shorter than visible light so is in the ultraviolet.
hc
(b) E =
= 1.03 × 1018 J = 6.44 eV

λ

(c) P =

39.39.

Etot NE
Pt (1.50 × 10−3 W)(12.0 × 10−9 s)
=
so N =
=
= 1.75 × 107 photons
t
t
E
1.03 × 10−18 J

EVALUATE: A very small amount of energy is delivered to the lens in each pulse, but this still
corresponds to a large number of photons.
n
− ( E − E )/kT
IDENTIFY: Apply Eq. (39.18): 5s = e 5s 3 p
n3 p
SET UP: E5s = 20.66 eV and E3 p = 18.70 eV
EXECUTE: E5s − E3 p = 20.66 eV − 18.70 eV = 1.96 eV(1.602 × 10−19 J/1 eV) = 3.140 × 10−19 J

39.40.

(a)

−19
−23
n5s
= e− (3.140 ×10 J)/[(1.38×10 J/K)(300 K)] = e−75.79 = 1.2 × 10−33
n3 p

(b)

−19
−23
n5s
= e− (3.140×10 J)/[(1.38×10 J/K)(600 K)] = e−37.90 = 3.5 × 10−17
n3 p

(c)

−19
−23
n5s
= e− (3.140×10 J)/[(1.38×10 J/K)(1200 K)] = e−18.95 = 5.9 × 10−9
n3 p

(d) EVALUATE: At each of these temperatures the number of atoms in the 5s excited state, the initial state for
the transition that emits 632.8 nm radiation, is quite small. The ratio increases as the temperature increases.
IDENTIFY: Apply Eq. (39.18).

SET UP: The energy of each of these excited states above the ground state is hc /λ , where λ is the
wavelength of the photon emitted in the transition from the excited state to the ground state.
n2 P3/ 2
= e − ( E2 P 3/ 2 − E2 P1/ 2 )/KT. From the diagram
EXECUTE:
n2 P1/ 2

ΔE3/2 − g =
ΔE1/2 −g =

hc

λ1
hc

λ2

=
=

(6.626 × 10−34 J)(2.998 × 108 m/s)
5.890 × 10−7 m
(6.626 × 10−34 J)(2.998 × 108 m/s)
5.896 × 10−7 m

= 3.373 × 10−19 J.
= 3.369 × 10−19 J. So ΔE3/ 2 −1/2 =

3.373 × 10−19 J − 3.369 × 10−19 J = 4.00 × 10−22 J.
n2 P3/ 2

n2 P1/ 2

= e − (4.00×10

−22

J)/(1.38 ×10−23 J/K⋅500 K)

= 0.944. So more atoms are in the 2P1/ 2 state.

EVALUATE: At this temperature kT = 6.9 × 10−21 J. This is greater than the energy separation between the

39.41.

states, so an atom has almost equal probability for being in either state, with only a small preference for the
lower energy state.
IDENTIFY: Energy radiates at the rate H = AeσT 4 .
SET UP: The surface area of a cylinder of radius r and length l is A = 2π rl.
1/4

⎛ H ⎞
EXECUTE: (a) T = ⎜
⎟
⎝ Aeσ ⎠

1/ 4

⎛
⎞
100 W

= ⎜⎜
2
4 ⎟
−3
−8
⎟ .
⎝ 2π (0.20 × 10 m)(0.30 m)(0.26)(5.671 × 10 W/m ⋅ K ) ⎠

T = 2.06 × 103 K.

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Particles Behaving as Waves

39-13

(b) λ mT = 2.90 × 10−3 m ⋅ K; λ m = 1410 nm.
EVALUATE: (c) λ m is in the infrared. The incandescent bulb is not a very efficient source of visible light
39.42.

because much of the emitted radiation is in the infrared.
IDENTIFY: Apply Eq. (39.21) and c = f λ .
SET UP: T in kelvins gives λ in meters.
EXECUTE: (a) λ m =

2.90 × 10−3 m ⋅ K
c
= 0.966 mm, and f =

= 3.10 × 1011 Hz.
3.00 K
λm

(b) A factor of 100 increase in the temperature lowers λ m by a factor of 100 to 9.66 μ m and raises the

frequency by the same factor, to 3.10 × 1013 Hz.
(c) Similarly, λ m = 966 nm and f = 3.10 × 1014 Hz.
EVALUATE: λ m decreases when T increases, as explained in the textbook.
39.43.

IDENTIFY and SET UP: The wavelength λ m where the Planck distribution peaks is given by Eq. (39.21).

39.44.

2.90 × 10−3 m ⋅ K
= 1.06 × 10−3 m = 1.06 mm.
2.728 K
EVALUATE: This wavelength is in the microwave portion of the electromagnetic spectrum. This radiation
is often referred to as the “microwave background” (Section 44.7). Note that in Eq. (39.21), T must be in
kelvins.
IDENTIFY and SET UP: Apply Eq. (39.21).
EXECUTE: λ m =

EXECUTE: T =

2.90 × 10−3 m ⋅ K

λm

=

2.90 × 10−3 m ⋅ K
400 × 10−9 m

= 7.25 × 103 K.

EVALUATE: 400 nm = 0.4 μm. This is shorter than any of the λ m values shown in Figure 39.32 in the
39.45.

textbook, and the temperature is therefore higher than those in the figure.
IDENTIFY: Since the stars radiate as blackbodies, they obey the Stefan-Boltzmann law and Wien’s
displacement law.
SET UP: The Stefan-Boltzmann law says that the intensity of the radiation is I = σT 4, so the total

radiated power is P = σ AT 4. Wien’s displacement law tells us that the peak-intensity wavelength is
λm = (constant)/T .
EXECUTE: (a) The hot and cool stars radiate the same total power, so the Stefan-Boltzmann law gives
σ Ah Th4 = σ AcTc4 ⇒ 4π Rh2Th4 = 4π Rc2Tc4 = 4π (3Rh ) 2 Tc4 ⇒ Th4 = 9T 4 ⇒ Th = T 3 = 1.7T , rounded to two

significant digits.
(b) Using Wien’s law, we take the ratio of the wavelengths, giving

λ m (hot) Tc
T
1
=
=
=
= 0.58, rounded to two significant digits.

λ m (cool) Th T 3
3
39.46.

EVALUATE: Although the hot star has only1/9 the surface area of the cool star, its absolute temperature
has to be only 1.7 times as great to radiate the same amount of energy.
IDENTIFY: Since the stars radiate as blackbodies, they obey the Stefan-Boltzmann law.
SET UP: The Stefan-Boltzmann law says that the intensity of the radiation is I = σ T 4 , so the total

radiated power is P = σ AT 4.
EXECUTE: (a) I = σT 4 = (5.67 × 10−8 W/m 2 ⋅ K 4 )(24,000 K) 4 = 1.9 × 1010 W/m 2
(b) Wien’s law gives λm = (0.00290 m ⋅ K)/(24,000 K) = 1.2 × 10 –7 m = 20 nm

This is not visible since the wavelength is less than 400 nm.
(c) P = AI ⇒ 4π R 2 = P /I = (1.00 × 1025 W)/(1.9 × 1010 W/m 2 )
which gives RSirius = 6.51 × 106 m = 6510 km.

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39-14

Chapter 39

RSirius /Rsun = (6.51 × 106 m)/(6.96 × 109 m) = 0.0093, which gives
RSirius = 0.0093 Rsun ≈ 1% Rsun
(d) Using the Stefan-Boltzmann law, we have
2

39.47.

4

2

4

4
2
4
⎛ R
⎞ ⎛ T
⎞
⎛
⎞ ⎛ 5800 K ⎞
Psun
σ AsunTsun
4π Rsun
Tsun
P
Rsun
=
=
= ⎜ sun ⎟ ⎜ sun ⎟ ⋅ sun = ⎜
⎟ ⎜
⎟ = 39
4
2
4

PSirius σ ASiriusTSirius 4π RSiriusTSirius ⎜⎝ RSirius ⎟⎠ ⎜⎝ TSirius ⎟⎠ PSirius ⎝ 0.00935 Rsun ⎠ ⎝ 24,000 K ⎠
EVALUATE: Even though the absolute surface temperature of Sirius B is about 4 times that of our sun, it
radiates only 1/39 times as much energy per second as our sun because it is so small.
IDENTIFY: Apply the Wien displacement law to relate λ m and T. Apply the Stefan-Boltzmann law to

relate the power output of the star to its surface area and therefore to its radius.
SET UP: For a sphere A = 4π r 2 . Since we assume a blackbody, e = 1.

2.90 × 10−3 K ⋅ m
k
= 9.7 × 10−8 m = 97 nm. This peak is in the
. λm =
30,000 K
T
ultraviolet region, which is not visible. The star is blue because the largest part of the visible light radiated
is in the blue/violet part of the visible spectrum.
(b) P = σ AT 4 (Stefan-Boltzmann law)
W ⎞
⎛
(100,000)(3.86 × 1026 W) = ⎜ 5.67 × 10−8 2 4 ⎟ (4π R 2 )(30,000 K)4
m K ⎠
⎝
EXECUTE: (a) Wien’s law: λm =

R = 8.2 × 109 m
Rstar /Rsun =

39.48.

8.2 × 109 m

6.96 × 108 m

= 12

EVALUATE: (c) The visual luminosity is proportional to the power radiated at visible wavelengths. Much
of the power is radiated nonvisible wavelengths, which does not contribute to the visible luminosity.
IDENTIFY: Since we know only that the mosquito is somewhere in the room, there is an uncertainty in its
position. The Heisenberg uncertainty principle tells us that there is an uncertainty in its momentum.
SET UP: The uncertainty principle is Δ xΔp x ≥ /2.
EXECUTE: (a) You know the mosquito is somewhere in the room, so the maximum uncertainty in its
horizontal position is Δx = 5.0 m.
(b) The uncertainty principle gives Δ xΔp x ≥ /2, and Δp x = mΔvx since we know the mosquito’s mass.

This gives Δ x mΔvx ≥ /2, which we can solve for Δvx to get the minimum uncertainty in vx .
Δv x =

39.49.

2mΔx

=

1.055 × 10−34 J ⋅ s
2(1.5 × 10−6 kg)(5.0 m)

= 7.0 × 10−30 m/s, which is hardly a serious impediment!

EVALUATE: For something as “large” as a mosquito, the uncertainty principle places a negligible
limitation on our ability to measure its speed.
(a) IDENTIFY and SET UP: Use Δ xΔpx ≥ /2 to calculate Δpx and obtain Δvx from this.

EXECUTE: Δpx ≥

Δv x =

2Δ x

=

1.055 × 10−34 J ⋅ s
2(1.00 × 10−6 m)

= 5.725 × 10−29 kg ⋅ m/s.

Δp x 5.275 × 10−29 kg ⋅ m/s
=
= 4.40 × 10−32 m/s.
m
1200 kg

(b) EVALUATE: Even for this very small Δ x the minimum Δvx required by the Heisenberg uncertainty

principle is very small. The uncertainty principle does not impose any practical limit on the simultaneous
measurements of the positions and velocities of ordinary objects.

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Particles Behaving as Waves
39.50.

39-15

IDENTIFY: Since we know that the marble is somewhere on the table, there is an uncertainty in its
position. The Heisenberg uncertainty principle tells us that there is therefore an uncertainty in its
momentum.
SET UP: The uncertainty principle is Δ xΔp x ≥ /2.
EXECUTE: (a) Since the marble is somewhere on the table, the maximum uncertainty in its horizontal
position is Δx = 1.75 m.
(b) Following the same procedure as in part (b) of Problem 39.48, the minimum uncertainty in the
1.055 × 10−34 J ⋅ s
horizontal velocity of the marble is Δvx =
=
= 3.01 × 10−33 m/s.
2mΔx 2(0.0100 kg)(1.75 m)
(c) The uncertainty principle tells us that we cannot know that the marble’s horizontal velocity is exactly
zero, so the smallest we could measure it to be is 3.01 × 10−33 m/s, from part (b). The longest time it could

remain on the table is the time to travel the full width of the table (1.75 m), so t = x /vx = (1.75 m)/
(3.01 × 10−33 m/s) = 5.81 × 1032 s = 1.84 × 1025 years. Since the universe is about 14 × 109 years old, this

39.51.

1.8 × 1025 yr

≈ 1.3 × 1015 times the age of the universe! Don’t hold your breath!
14 × 109 yr
EVALUATE: For household objects, the uncertainty principle places a negligible limitation on our ability
to measure their speed.
IDENTIFY: Heisenberg’s Uncertainty Principles tells us that Δ xΔp x ≥ /2.

time is about

SET UP: We can treat the standard deviation as a direct measure of uncertainty.
EXECUTE: Here Δ xΔp x = (1.2 × 10−10 m)(3.0 × 10−25 kg ⋅ m/s) = 3.6 × 10−35 J ⋅ s, but

/2 = 5.28 × 10−35 J ⋅ s. Therefore Δ xΔp x < /2, so the claim is not valid .
EVALUATE: The uncertainty product ΔxΔp x must increase by a factor of about 1.5 to become consistent
39.52.

with the Heisenberg Uncertainty Principle.
IDENTIFY: Apply the Heisenberg Uncertainty Principle.
SET UP: Δpx = mΔvx .
EXECUTE: (a) (Δ x)(mΔvx ) ≥ /2, and setting Δvx = (0.010)vx and the product of the uncertainties equal

to /2 (for the minimum uncertainty) gives vx = /[2m(0.010)Δ x] = 29.0 m/s.
(b) Repeating with the proton mass gives 15.8 mm/s.
EVALUATE: For a given Δpx , Δvx is smaller for a proton than for an electron, since the proton has larger
39.53.

mass.
IDENTIFY: Apply the Heisenberg Uncertainty Principle in the form ΔE Δt = /2.
SET UP: Let Δt = 5.2 × 10−3 s, the lifetime of the state of the atom, and let ΔE be the uncertainty in the

energy of the state.
EXECUTE: ΔE >

39.54.

2Δt

=

(1.055 × 10−34 J ⋅ s)
−3

= 1.01 × 10−32 J = 6.34 × 10−14 eV.

2(5.2 × 10 s)
EVALUATE: The uncertainty in the energy is a very small fraction of the typical energy of atomic states,
which is on the order of 1 eV.
IDENTIFY and SET UP: The Heisenberg Uncertainty Principle says Δ xΔp x ≥ ប /2. The minimum allowed
Δ xΔp x is /2. Δpx = mΔvx .
EXECUTE: (a) mΔ xΔvx = /2. Δvx =
(b) Δ x =

2mΔvx

=

1.055 × 10

−34

2mΔ x

=

1.055 × 10−34 J ⋅ s
2(1.67 × 10

J⋅s

2(9.11 × 10−31 kg)(0.250 m/s)

−27

kg)(2.0 × 10

−12

m)

= 1.6 × 104 m/s.

= 2.3 × 10−4 m.

EVALUATE: The smaller Δx is, the larger Δvx must be.

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39-16

39.55.

Chapter 39

(a) IDENTIFY and SET UP: Apply Eq. (39.17): mr =
EXECUTE: mr =

207 me mp
m1m2
=
m1 + m2 207 me + mp

207(9.109 × 10−31 kg)(1.673 × 10−27 kg)
207(9.109 × 10−31 kg) + 1.673 × 10−27 kg

= 1.69 × 10−28 kg

We have used me to denote the electron mass.
(b) IDENTIFY: In Eq. (39.14) replace m = me by mr : En = −

1 mr e 4

⑀ 02 8n 2h 2

.

⎛ m ⎞ ⎛ 1 m e4 ⎞
1 m e4
SET UP: Write as En = ⎜ r ⎟ ⎜ − 2 H2 2 ⎟ , since we know that 2 H 2 = 13.60 eV. Here mH denotes
⎜
⎟
⑀ 0 8h
⎝ mH ⎠ ⎝ ⑀ 0 8n h ⎠

the reduced mass for the hydrogen atom; mH = 0.99946(9.109 × 10−31 kg) = 9.104 × 10−31 kg.
⎛ m ⎞ ⎛ 13.60 eV ⎞

EXECUTE: En = ⎜ r ⎟ ⎜ −
⎟
n2
⎠
⎝ mH ⎠ ⎝

E1 =

1.69 × 10−28 kg
9.109 × 10−31 kg

(−13.60 eV) = 186(−13.60 eV) = −2.53 keV

⎛ m ⎞ ⎛ R ch ⎞
(c) SET UP: From part (b), En = ⎜ r ⎟ ⎜ − H2 ⎟ , where RH = 1.097 × 107 m −1 is the Rydberg constant
⎝ mH ⎠ ⎝ n ⎠
hc
= Ei − Ef to find an expression for 1/λ . The initial level for
for the hydrogen atom. Use this result in

λ

the transition is the ni = 2 level and the final level is the nf = 1 level.
EXECUTE:

1

λ
1

=
=

hc

λ

=

mr ⎛ RH ch ⎛ RH ch ⎞ ⎞
− ⎜ − 2 ⎟⎟
⎜−
⎜ n ⎟⎟
mH ⎜⎝ ni2
⎝
f ⎠⎠

⎛ 1
mr
1 ⎞
RH ⎜ 2 − 2 ⎟
⎜
⎟
mH
⎝ nf ni ⎠

1.69 × 10−28 kg

λ 9.109 × 10
λ = 0.655 nm

−31

⎛1 1 ⎞
(1.097 × 107 m −1) ⎜ 2 − 2 ⎟ = 1.527 × 109 m −1
kg
⎝1 2 ⎠

EVALUATE: From Example 39.6 the wavelength of the radiation emitted in this transition in hydrogen is
m
122 nm. The wavelength for muonium is H = 5.39 × 10−3 times this. The reduced mass for hydrogen is
mr
very close to the electron mass because the electron mass is much less then the proton mass:
mp /me = 1836. The muon mass is 207 me = 1.886 × 10−28 kg. The proton is only about 10 times more

massive than the muon, so the reduced mass is somewhat smaller than the muon mass. The muon-proton
atom has much more strongly bound energy levels and much shorter wavelengths in its spectrum than for
hydrogen.
39.56.

IDENTIFY: Apply conservation of momentum to the system of atom and emitted photon.
SET UP: Assume the atom is initially at rest. For a photon E =

hc

λ

and p =

h

λ

.

EXECUTE: (a) Assume a non-relativistic velocity and conserve momentum ⇒ mv =
2

h

λ

⇒v=

h
.
mλ

2

1
1 ⎛ h ⎞
h
(b) K = mv 2 = m ⎜
.
⎟ =
2
2 ⎝ mλ ⎠
2mλ 2

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Particles Behaving as Waves

39-17

K
h2
λ
h
=
⋅
=
. Recoil becomes an important concern for small m and small λ since this
E 2mλ 2 hc 2mcλ
ratio becomes large in those limits.
hc (6.63 × 10−34 J ⋅ s)(3.00 × 108 m/s)
(d) E = 10.2 eV ⇒ λ =
=
= 1.22 × 10−7 m = 122 nm.
E
(10.2 eV)(1.60 × 10−19 J/eV)
(c)

K=

39.57.

(6.63 × 10−34 J ⋅ s) 2
2(1.67 × 10−27 kg)(1.22 × 10−7 m) 2

= 8.84 × 10−27 J = 5.53 × 10−8 eV.

K 5.53 × 10−8 eV
=
= 5.42 × 10−9. This is quite small so recoil can be neglected.
E
10.2 eV
EVALUATE: For emission of photons with ultraviolet or longer wavelengths the recoil kinetic energy of
the atom is much less than the energy of the emitted photon.
IDENTIFY and SET UP: The Hα line in the Balmer series corresponds to the n = 3 to n = 2 transition.
En = −

13.6 eV
n2

.

hc

λ

= ΔE.

⎛ 1 1⎞
EXECUTE: (a) The atom must be given an amount of energy E3 − E1 = −(13.6 eV) ⎜ 2 − 2 ⎟ = 12.1 eV.
⎝3 1 ⎠
hc

(b) There are three possible transitions. n = 3 → n = 1: ΔE = 12.1 eV and λ =
= 103 nm;
ΔE
1 ⎞
⎛ 1
n = 3 → n = 2 : ΔE = −(13.6 eV) ⎜ 2 − 2 ⎟ = 1.89 eV and λ = 657 nm; n = 2 → n = 1:
2 ⎠
⎝3

39.58.

⎛ 1 1⎞
ΔE = −(13.6 eV) ⎜ 2 − 2 ⎟ = 10.2 eV and λ = 122 nm.
⎝2 1 ⎠
EVALUATE: The larger the transition energy for the atom, the shorter the wavelength.
n
− ( E − E )/kT
IDENTIFY: Apply 2 = e ex g .
n1

−13.6 eV
= −3.4 eV. Eg = −13.6 eV. Eex − Eg = 10.2 eV = 1.63 × 10−18 J.
4
−( Eex − Eg ) n2
− (1.63 × 10−18 J)
.
= 10−12. T =
= 4275 K.
EXECUTE: (a) T =
k ln( n2 /n1) n1

(1.38 × 10−23 J/K) ln(10−12 )
SET UP: Eex = E2 =

(b)

− (1.63 × 10−18 J)
n2
= 6412 K.
= 10−8. T =
n1
(1.38 × 10−23 J/K) ln(10−8 )

n2
− (1.63 × 10−18 J)
= 10−4. T =
= 12824 K.
n1
(1.38 × 10−23 J/K) ln(10−4 )
EVALUATE: (d) For absorption to take place in the Balmer series, hydrogen must start in the n = 2 state.
From part (a), colder stars have fewer atoms in this state leading to weaker absorption lines.
(a) IDENTIFY and SET UP: The photon energy is given to the electron in the atom. Some of this energy
overcomes the binding energy of the atom and what is left appears as kinetic energy of the free electron.
Apply hf = Ef − Ei , the energy given to the electron in the atom when a photon is absorbed.
(c)

39.59.

EXECUTE: The energy of one photon is

hc

λ

hc

λ

=

(6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s)
85.5 × 10−9 m

= 2.323 × 10−18 J(1 eV/1.602 × 10−19 J) = 14.50 eV.

The final energy of the electron is Ef = Ei + hf . In the ground state of the hydrogen atom the energy of the
electron is Ei = −13.60 eV. Thus Ef = −13.60 eV + 14.50 eV = 0.90 eV.

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39-18

Chapter 39
(b) EVALUATE: At thermal equilibrium a few atoms will be in the n = 2 excited levels, which have an
energy of −13.6 eV/4 = −3.40 eV,10.2 eV greater than the energy of the ground state. If an electron with
E = −3.40 eV gains 14.5 eV from the absorbed photon, it will end up with 14.5 eV − 3.4 eV = 11.1 eV of

kinetic energy.
39.60.

IDENTIFY: For circular motion, L = mvr and a =

v2
mM
. Newton’s law of gravitation is Fg = G 2 ,
r
r

with G = 6.67 × 10−11 N ⋅ m 2 /kg 2 .
SET UP: The period T is 2.00 h = 7200 s.
EXECUTE: (a) mvr = n

n=

h
2π mvr
2π r
.n =
.v =
So
h
T
2π

(2π r )2 m (2π) 2 (8.06 × 106 m) 2 (20.0 kg)
=
= 1.08 × 1046.
hT
(6.63 × 10−34 J . s)(7200 s)

(b) F = ma gives G

mmE
r2

=m

v 2 GmE
nh
GmE
n2h2
.
= v 2 . The Bohr postulate says v =
so
= 2 2 2
r
r
2πmr
r
4π m r

⎛
⎞ 2
h2
r =⎜ 2
n . This is in the form r = kn 2, with
⎜ 4π Gm m 2 ⎟⎟
E
⎝

⎠

h2

k=

2

4π GmE m

2

=

(6.63 × 10−34 J.s)2
2

4π (6.67 × 10

−11

N ⋅ m 2 /kg 2 )(5.97 × 1024 kg) 2

= 7.0 × 10−86 m

(c) Δr = rn +1 − rn = k ([n + 1]2 − n 2 ) = (2n + 1)k = (2[1.08 × 1046 ] + 1)(7.0 × 10−86 m) = 1.5 × 10−39 m
EVALUATE: (d) Δr is exceedingly small, so the separation of adjacent orbits is not observable.
(e) There is no measurable difference between quantized and classical orbits for this satellite; either
method of calculation is totally acceptable.
39.61.

IDENTIFY: Assuming that Betelgeuse radiates like a perfect blackbody, Wien’s displacement and the
Stefan-Boltzmann law apply to its radiation.
SET UP: Wien’s displacement law is λ peak =

2.90 × 10−3 m ⋅ K
, and the Stefan-Boltzmann law says that
T

the intensity of the radiation is I = σ T 4 , so the total radiated power is P = σ AT 4 .
EXECUTE: (a) First use Wien’s law to find the peak wavelength:

λ m = (2.90 × 10−3 m ⋅ K)/(3000 K) = 9.667 × 10−7 m
Call N the number of photons/second radiated. N × (energy per photon) = IA = σ AT 4 .

N (hc/λm ) = σ AT 4 . N =
N=

λmσ AT 4
hc

.

(9.667 × 10−7 m)(5.67 × 10−8 W/m 2 ⋅ K 4 )(4π )(600 × 6.96 × 108 m) 2 (3000 K) 4
(6.626 × 10−34 J ⋅ s)(3.00 × 108 m/s)

.

N = 5 × 1049 photons/s.
2

(b)

4

I B AB σ ABTB4 4π RB2TB4 ⎛ 600 RS ⎞ ⎛ 3000 K ⎞
4
=
=
=⎜
⎟ ⎜
⎟ = 3 × 10
IS AS σ ASTS4 4π RS2TS4 ⎝ RS ⎠ ⎝ 5800 K ⎠

EVALUATE: Betelgeuse radiates 30,000 times as much energy per second as does our sun!

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Particles Behaving as Waves
39.62.

39-19

IDENTIFY: The diffraction grating allows us to determine the peak-intensity wavelength of the light. Then
Wien’s displacement law allows us to calculate the temperature of the blackbody, and the StefanBoltzmann law allows us to calculate the rate at which it radiates energy.
SET UP: The bright spots for a diffraction grating occur when d sin θ = mλ . Wien’s displacement law is

λ peak =

2.90 × 10−3 m ⋅ K
, and the Stefan-Boltzmann law says that the intensity of the radiation is
T

I = σ T 4 , so the total radiated power is P = σ AT 4 .

EXECUTE: (a) First find the wavelength of the light:

λ = d sin θ = [1/(385,000 lines/m)] sin(11.6°) = 5.22 × 10−7 m
Now use Wien’s law to find the temperature: T = (2.90 × 10−3 m ⋅ K)/(5.22 × 10−7 m) = 5550 K.
(b) The energy radiated by the blackbody is equal to the power times the time, giving
U = Pt = IAt = σ AT 4t , which gives
t = U /(σ AT 4 ) = (12.0 × 106 J)/[(5.67 × 10−8 W/m 2 ⋅ K 4 )(4π )(0.0750 m)2 (5550 K)4 ] = 3.16 s.

39.63.

EVALUATE: By ordinary standards, this blackbody is very hot, so it does not take long to radiate 12.0 MJ
of energy.
IDENTIFY: The energy of the peak-intensity photons must be equal to the energy difference between the
n = 1 and the n = 4 states. Wien’s law allows us to calculate what the temperature of the blackbody must
be for it to radiate with its peak intensity at this wavelength.
13.6 eV
SET UP: In the Bohr model, the energy of an electron in shell n is En = −
, and Wien’s
n2

2.90 × 10−3 m ⋅ K
. The energy of a photon is E = hf = hc /λ .
T

EXECUTE: First find the energy (ΔE) that a photon would need to excite the atom. The ground state of
the atom is n = 1 and the third excited state is n = 4. This energy is the difference between the two energy
⎛ 1 1⎞
levels. Therefore ΔE = (−13.6 eV) ⎜ 2 − 2 ⎟ = 12.8 eV. Now find the wavelength of the photon having
⎝4 1 ⎠
this amount of energy. hc /λ = 12.8 eV and

displacement law is λ m =

λ = (4.136 × 10−15 eV ⋅ s)(3.00 × 108 m/s)/(12.8 eV) = 9.73 × 10−8 m
Now use Wien’s law to find the temperature. T = (0.00290 m ⋅ K)/(9.73 × 10−8 m) = 2.98 × 104 K.

39.64.

EVALUATE: This temperature is well above ordinary room temperatures, which is why hydrogen atoms
are not in excited states during everyday conditions.
IDENTIFY: The blackbody radiates heat into the water, but the water also radiates heat back into the
blackbody. The net heat entering the water causes evaporation. Wien’s law tells us the peak wavelength
radiated, but a thermophile in the water measures the wavelength and frequency of the light in the water.
SET UP: By the Stefan-Boltzman law, the net power radiated by the blackbody is
dQ
4
4
= σ A Tsphere
− Twater
. Since this heat evaporates water, the rate at which water evaporates is
dt

(

)

dQ
dm
2.90 × 10−3 m ⋅ K
= Lv
. Wien’s displacement law is λ m =
, and the wavelength in the water is
dt
dt
T
λ w =λ0 /n.
EXECUTE: (a) The net radiated heat is

(

)

dQ
4
4
= σ A Tsphere
− Twater
and the evaporation rate is
dt

dQ
dm
= Lv
, where dm is the mass of water that evaporates in time dt. Equating these two rates gives

dt
dt

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39-20

Chapter 39

(

)

2
4
4
dm
dm σ (4π R ) Tsphere − Twater
4
4
Lv
= σ A Tsphere − Twater .
=
.
dt
dt
Lv

(

dm (5.67 × 10
=
dt

)

−8

W/m 2 ⋅ K 4 )(4π )(0.120 m) 2 ⎡ (498 K) 4 − (373 K) 4 ⎤
⎣
⎦ = 1.92 × 10−4 kg/s = 0.193 g/s
2256 × 103 J/kg

(b) (i) Wien’s law gives λ m = (0.00290 m ⋅ K)/(498 K) = 5.82 × 10−6 m

But this would be the wavelength in vacuum. In the water the thermophile organism would measure
λ w = λ0 /n = (5.82 × 10−6 m)/1.333 = 4.37 × 10−6 m = 4.37 µm
(ii) The frequency is the same as if the wave were in air, so
f = c /λ0 = (3.00 × 108 m/s)/(5.82 × 10−6 m) = 5.15 × 1013 Hz
EVALUATE: An alternative way is to use the quantities in the water: f =

39.65.

c /n

λ0 /n

= c /λ0 , which gives the

same answer for the frequency. An organism in the water would measure the light coming to it through the
water, so the wavelength it would measure would be reduced by a factor of 1/n.
IDENTIFY: Apply conservation of energy and conservation of linear momentum to the system of atom
plus photon.
(a) SET UP: Let Etr be the transition energy, Eph be the energy of the photon with wavelength λ ′, and
Er be the kinetic energy of the recoiling atom. Conservation of energy gives Eph + Er = Etr .
Eph =

hc
hc
hc
so
= Etr − Er and λ ′ =
.
λ′
λ′
Etr − Er

EXECUTE: If the recoil energy is neglected then the photon wavelength is λ = hc /Etr .

⎛
⎞
1
1 ⎞ ⎛ hc ⎞⎛
1
Δλ = λ ′ − λ = hc ⎜
−
− 1⎟
⎟=⎜

⎟⎜
−
−
E
E
E
E
1
E
/
E
r
tr ⎠ ⎝ tr ⎠⎝
r tr
⎝ tr
⎠
−1

⎛
E
1
E ⎞
E
1
= ⎜1 − r ⎟ ≈ 1 + r since r
Etr
Etr
1 − Er /Etr ⎝ Etr ⎠
(We have used the binomial theorem, Appendix B.)

Thus Δλ =

hc ⎛ Er ⎞
⎛ Er ⎞ 2
⎜
⎟ , or since Etr = hc /λ , Δλ = ⎜ ⎟ λ .
Etr ⎝ Etr ⎠
⎝ hc ⎠

SET UP: Use conservation of linear momentum to find Er : Assuming that the atom is initially at rest, the

momentum pr of the recoiling atom must be equal in magnitude and opposite in direction to the
momentum pph = h /λ of the emitted photon: h /λ = pr .
EXECUTE: Er =

h2
pr2
.
, where m is the mass of the atom, so Er =
2m
2mλ 2

⎛ h 2 ⎞⎛ λ 2 ⎞
h
⎛E ⎞
Use this result in the above equation: Δλ = ⎜ r ⎟ λ 2 = ⎜
;
⎟⎟ =
2 ⎟⎜
⎜

⎟⎜
hc
hc
mc
2
m
λ
2
⎝ ⎠
⎝
⎠⎝ ⎠
note that this result for Δλ is independent of the atomic transition energy.
(b) For a hydrogen atom m = mp and Δλ =

6.626 × 10−34 J ⋅ s
h
=
= 6.61 × 10−16 m
2mpc 2(1.673 × 10−27 kg)(2.998 × 108 m/s)

EVALUATE: The correction is independent of n. The wavelengths of photons emitted in hydrogen atom
transitions are on the order of 100 nm = 10−7 m, so the recoil correction is exceedingly small.

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Particles Behaving as Waves
39.66.

39-21

IDENTIFY: Combine I = σT 4 , P = IA, and ΔE = Pt.
SET UP: In the Stefan-Boltzmann law the temperature must be in kelvins. 200°C = 473 K.

ΔE

(100 J)

= 8.81 × 103 s = 2.45 h.
(4.00 × 10 m )(5.67 × 10−8 W/m 2 ⋅ K 4 )(473 K)4
Aσ T
EVALUATE: P = 0.0114 W. Since the area of the hole is small, the rate at which the cavity radiates
EXECUTE: t =

39.67.

4

=

−6

2

energy through the hole is very small.
IDENTIFY and SET UP: Follow the procedures specified in the problem.
2π hc 2
c
2π hc 2

2π hf 5
EXECUTE: (a) I (λ ) = 5 hc /λ kT
=
but λ = ⇒ I ( f ) =
f
− 1)
λ (e
(c /f )5 (ehf /kT − 1) c3 (ehf /kT − 1)
(b)

∫0

⎛ −c ⎞
0
I (λ ) d λ = ∫ I ( f ) df ⎜⎜ 2 ⎟⎟
∞
⎝ f ⎠

=∫

∞

2π hf 3 df

∞

0

c 2 (ehf /kT − 1)

(c) The expression

=

2π (kT ) 4
c 2 h3

2π 5k 4
15h3c 2

∞

∫0

x3
ex − 1

dx =

2π ( kT ) 4 1
(2π )5 (kT ) 4 2π 5k 4T 4
4
=
=
(2
π
)
.
240h3c 2
15c 2h3

c 2h3 240

= σ as shown in Eq. (39.28). Plugging in the values for the constants we get

σ = 5.67 × 10−8 W/m 2 ⋅ K 4 .
39.68.

EVALUATE: The Planck radiation law, Eq. (39.24), predicts the Stefan-Boltzmann law, Eq. (39.19).
h
h
IDENTIFY: λ = =
. From Chapter 36, if λ a then the width w of the central maximum is
p
2mE
Rλ
w=2
, where R = 2.5 m and a is the width of the slit.
a
SET UP: vx =

2E
, since the beam is traveling in the x-direction and Δv y
m

EXECUTE: (a) λ =
(b)

vx

(6.63 × 10−34 J ⋅ s)

h
=
= 1.94 × 10−10 m.
−31
−19
2mK
2(9.11 × 10 kg)(40 eV)(1.60 × 10 J/eV)

(2.5 m)(9.11 × 10−31 kg)1/2
R
R
=
=
= 6.67 × 10−7 s.
−19
v
2 E /m
2(40 eV)(1.6 × 10 J/eV)

λ

(c) The width w is w = 2 R ' and w = Δv yt = Δp yt /m, where t is the time found in part (b) and a is the slit
a
2mλ R
width. Combining the expressions for w, Δp y =
= 2.65 × 10−28 kg ⋅ m/s.
at
(d) Δy =

2 Δp y

= 0.20 μm, which is the same order of magnitude of the width of the slit.

EVALUATE: For these electrons λ = 1.94 × 10−10 m. This is much smaller than a and the approximate

Δp y
2Rλ
2E
is very accurate. Also, vx =
= 2.9 × 102 m/s, so it
= 3.75 × 106 m/s. Δv y =
a
m
m
is the case that vx Δv y .

expression w =

39.69.

p2
= qΔV , where ΔV is the
2m
λ
λ
accelerating voltage. To exhibit wave nature when passing through an opening, the de Broglie wavelength
of the particle must be comparable with the width of the opening.
SET UP: An electron has mass 9.109 × 10−31 kg. A proton has mass 1.673 × 10−27 kg.
EXECUTE: (a) E = hc /λ = 12 eV
IDENTIFY: For a photon E =

hc

. For a particle with mass, p =

h

and E =

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39-22

Chapter 39
(b) Find E for an electron with λ = 0.10 × 10−6 m. λ = h /p so p = h /λ = 6.626 × 10−27 kg ⋅ m/s.

E = p 2 /(2m) = 1.5 × 10−4 eV. E = qΔV so ΔV = 1.5 × 10−4 V.
v = p /m = (6.626 × 10−27 kg ⋅ m/s)/(9.109 × 10−31 kg) = 7.3 × 103 m/s

(c) Same λ so same p. E = p 2 /(2m) but now m = 1.673 × 10−27 kg so E = 8.2 × 10−8 eV and

ΔV = 8.2 × 10−8 V. v = p /m = (6.626 × 10−27 kg ⋅ m/s)/(1.673 × 10−27 kg) = 4.0 m/s

39.70.

EVALUATE: A proton must be traveling much slower than an electron in order to have the same de
Broglie wavelength.
IDENTIFY: The de Broglie wavelength of the electrons must be such that the first diffraction minimum

occurs at θ = 20.0°.
h
SET UP: The single-slit diffraction minima occur at angles θ given by a sin θ = mλ. p = .

λ

EXECUTE: (a) λ = a sin θ = (150 × 10

v=

6.626 × 10

−34

J⋅s

(9.11 × 10−31 kg)(5.13 × 10−8 m)

−9

m)sin 20° = 5.13 × 10

−8

m. λ = h /mv → v = h /mλ .

= 1.42 × 104 m/s.

(b) No electrons strike the screen at the location of the second diffraction minimum. a sin θ 2 = 2λ.

⎛ 5.13 × 10−8 m ⎞
= ±2 ⎜
= ±0.684. θ 2 = ±43.2°.
⎜ 150 × 10−9 m ⎟⎟
a
⎝
⎠
EVALUATE: The intensity distribution in the diffraction pattern depends on the wavelength λ and is the
same for light of wavelength λ as for electrons with de Broglie wavelength λ .
IDENTIFY: The electrons behave like waves and produce a double-slit interference pattern after passing
through the slits.
SET UP: The first angle at which destructive interference occurs is given by d sin θ = λ /2. The de Broglie
wavelength of each of the electrons is λ = h /mv.
EXECUTE: (a) First find the wavelength of the electrons. For the first dark fringe, we have d sin θ = λ /2,
which gives (1.25 nm)(sin 18.0°) = λ /2, and λ = 0.7725 nm. Now solve the de Broglie wavelength

sin θ 2 = ±2

39.71.

λ

equation for the speed of the electron:
v=

6.626 × 10−34 J ⋅ s
h
=
= 9.42 × 105 m/s
mλ (9.11 × 10−31 kg)(0.7725 × 10−9 m)

which is about 0.3% the speed of light, so they are nonrelativistic.
(b) Energy conservation gives eV = 12 mv 2 and
V = mv 2 /2e = (9.11 × 10−31 kg)(9.42 × 105 m/s) 2 /[2(1.60 × 10−19 C)] = 2.52 V

EVALUATE: The hole must be much smaller than the wavelength of visible light for the electrons to show
diffraction.
39.72.

IDENTIFY: The alpha particles and protons behave as waves and exhibit circular-aperture diffraction after
passing through the hole.
SET UP: For a round hole, the first dark ring occurs at the angle θ for which sin θ = 1.22λ /D, where D is
the diameter of the hole. The de Broglie wavelength for a particle is λ = h /p = h /mv.
EXECUTE: Taking the ratio of the sines for the alpha particle and proton gives

sin θα 1.22λα λα
=
=
sin θ p 1.22λ p λ p
The de Broglie wavelength gives λ p = h /pp and λα = h /pα , so

sin θα h /pα pp
=
=
. Using K = p 2 /2m,
sin θ p h /pp pα

we have p = 2mK . Since the alpha particle has twice the charge of the proton and both are accelerated
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Particles Behaving as Waves

39-23

through the same potential difference, Kα = 2 K p . Therefore pp = 2mp K p and

pα = 2mα Kα = 2mα (2 K p ) = 4mα K p . Substituting these quantities into the ratio of the sines gives
2mp K p
mp
sin θα pp
=
=
=
sin θ p pα
2
mα
4mα K p
Solving for sin θα gives sin θα =

1.67 × 10−27 kg
2(6.64 × 10−27 kg)

sin15.0° and θα = 5.3°.

EVALUATE: Since sin θ is inversely proportional to the mass of the particle, the larger-mass alpha
particles form their first dark ring at a smaller angle than the ring for the lighter protons.
39.73.

IDENTIFY: Both the electrons and photons behave like waves and exhibit single-slit diffraction after
passing through their respective slits.
SET UP: The energy of the photon is E = hc /λ and the de Broglie wavelength of the electron is
λ = h /mv = h /p. Destructive interference for a single slit first occurs when a sin θ = λ.
EXECUTE: (a) For the photon: λ = hc /E and a sin θ = λ . Since the a and θ are the same for the photons
and electrons, they must both have the same wavelength. Equating these two expressions for λ gives
h
a sinθ = hc /E. For the electron, λ = h /p =
and a sinθ = λ . Equating these two expressions for λ
2mK
h
h
. Equating the two expressions for a sin θ gives hc /E =
, which
gives a sin θ =
2mK
2mK

gives E = c 2mK = (4.05 × 10−7 J1/ 2 ) K .
E c 2mK
2mc 2
=
=
. Since v
c, mc 2 > K , so the square root is > 1. Therefore E /K > 1,
K
K
K
meaning that the photon has more energy than the electron.
EVALUATE: When a photon and a particle have the same wavelength, the photon has more energy than

the particle.
IDENTIFY: The de Broglie wavelength of the electrons must equal the wavelength of the light.
SET UP: The maxima in the two-slit interference pattern are located by d sin θ = mλ . For an electron,
h
h
λ= =
.
p mv
(b)

39.74.

EXECUTE: λ =

d sin θ (40.0 × 10−6 m)sin(0.0300 rad)
=
= 600 nm. The velocity of an electron with this
m
2

(6.63 × 10−34 J ⋅ s)
p
h
=
=
= 1.21 × 103 m/s. Since
m mλ (9.11 × 10−31 kg)(600 × 10−9 m)
this velocity is much smaller than c we can calculate the energy of the electron classically
1
1

K = mv 2 = (9.11 × 10−31 kg)(1.21 × 103 m/s)2 = 6.70 × 10−25 J = 4.19 μeV.
2
2
hc
EVALUATE: The energy of the photons of this wavelength is E =
= 2.07 eV. The photons and
wavelength is given by Eq. (39.1). v =

λ

electrons have the same wavelength but very different energies.
39.75.

IDENTIFY and SET UP: The de Broglie wavelength of the blood cell is λ =

h
.
mv

6.63 × 10−34 J ⋅ s

= 1.66 × 10−17 m.
(1.00 × 10−14 kg)(4.00 × 10−3 m/s)
EVALUATE: We need not be concerned about wave behavior.
EXECUTE: λ =

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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

39-24
39.76.

Chapter 39
IDENTIFY: An electron and a photon both have the same wavelength. We want to use this fact to calculate
the energy of each of them.
h
SET UP: The de Broglie wavelength is λ = . The energy of the electron is its kinetic energy,
p

K = 12 mv 2 = p 2 /2m. The energy of the photon is E = hf = hc /λ .
EXECUTE: (a) p =

E=

λ

=

6.626 × 10−34 J ⋅ s
400 × 10−9 m

= 1.656 × 10−27 kg ⋅ m/s.

p 2 (1.656 × 10−27 kg ⋅ m/s) 2
=
= 1.506 × 10−24 J = 9.40 × 10−6 eV
2m
2(9.109 × 10−31 kg)
hc

(6.626 × 10−34 J ⋅ s)(2.998 × 108 m/s)

= 4.966 × 10−19 J = 3.10 eV
400 × 10−9 m
EVALUATE: The photon has around 300,000 times as much energy as the electron.
IDENTIFY and SET UP: Follow the procedures specified in the problem.
(b) E =

39.77.

h

λ

=

1/ 2

⎛ v2 ⎞
h ⎜1 − 2 ⎟
⎜ c ⎟
h
⎠
EXECUTE: (a) λ = = ⎝
p
mv

⇒ v2 =

(b) v =

39.78.

h2

⎛ 2 2 h ⎞
⎜⎜ λ m + 2 ⎟⎟
c ⎠
⎝
2

c
1/2

⎛ ⎛ λ ⎞2 ⎞
⎜1 + ⎜
⎟
⎜ ⎝ (h /mc ) ⎟⎠ ⎟
⎝
⎠

(c) λ = 1.00 × 10−15 m

⎛ v2 ⎞
h 2v 2
v2
⇒ λ 2 m 2v 2 = h 2 ⎜ 1 − 2 ⎟ = h 2 − 2 ⇒ λ 2 m 2v 2 + h 2 2 = h 2
⎜ c ⎟
c

c
⎝
⎠

=

c2

⎛λ m c
⎞
+ 1⎟
⎜⎜
2
⎟
h
⎝
⎠
2

2 2

⇒v=

c
1/ 2

⎛ ⎛ mcλ ⎞2 ⎞
⎜1 + ⎜
⎟
⎜ ⎝ h ⎟⎠ ⎟

⎝
⎠

.

⎛ 1 ⎛ mcλ ⎞2 ⎞
m 2c 2λ 2
⎟ = (1 − Δ)c. Δ =
.
≈ c ⎜1 − ⎜
⎟
⎜ 2⎝ h ⎠ ⎟
2h 2
⎝
⎠

h
(9.11 × 10−31 kg) 2 (3.00 × 108 m/s) 2 (1.00 × 10−15 m) 2
. Δ=
= 8.50 × 10−8
mc
2(6.63 × 10−34 J ⋅ s) 2

⇒ v = (1 − Δ )c = (1 − 8.50 × 10−8 )c.
EVALUATE: As Δ → 0, v → c and λ → 0.
IDENTIFY and SET UP: The minimum uncertainty product is Δ xΔp x = /2. Δ x = r1, where r1 is the

radius of the n = 1 Bohr orbit. In the n = 1 Bohr orbit, mv1r1 =

h

h
and p1 = mv1 =
.
2π
2π r1

1.055 × 10−34 J ⋅ s

= 1.0 × 10−24 kg ⋅ m/s. This is the same as the
2(0.529 × 10−10 m)
magnitude of the momentum of the electron in the n = 1 Bohr orbit.
EVALUATE: Since the momentum is the same order of magnitude as the uncertainty in the momentum,
the uncertainty principle plays a large role in the structure of atoms.
EXECUTE: Δpx =

39.79.

2Δ x

=

2r1

=

IDENTIFY and SET UP: Combining the two equations in the hint gives pc = K ( K + 2mc 2 ) and

λ=

hc

K ( K + 2mc 2 )

.

EXECUTE: (a) With K = 3mc 2 this becomes λ =
2

(b) (i) K = 3mc = 3(9.109 × 10

−31

hc
2

2

2

3mc (3mc + 2mc )

=

h
.
15mc

8

kg)(2.998 × 10 m/s) 2 = 2.456 × 1013 J = 1.53 MeV

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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


Particles Behaving as Waves

λ=

39-25

h
6.626 × 10−34 J ⋅ s
=
= 6.26 × 10−13 m
15mc
15(9.109 × 10−31 kg)(2.998 × 108 m/s)

(ii) K is proportional to m, so for a proton K = (mp /me )(1.53 MeV) = 1836(1.53 MeV) = 2810 MeV

λ is proportional to 1/m, so for a proton
λ = (me /mp )(6.26 × 10−13 m) = (1/1836)(6.26 × 10−13 m) = 3.41 × 10−16 m.

39.80.

EVALUATE: The proton has a larger rest mass energy so its kinetic energy is larger when K = 3mc 2 .
The proton also has larger momentum so has a smaller λ .
IDENTIFY: Apply the Heisenberg Uncertainty Principle. Consider only one component of position and
momentum.
SET UP: Δ xΔp x ≥ /2. Take Δx ≈ 5.0 × 10−15 m. K = E − mc 2 . For a proton, m = 1.67 × 10−27 kg.
EXECUTE: (a) Δpx =

2 Δx

=

(1.055 × 10−34 J ⋅ s)
2(5.0 × 10−15 m)

= 1.1 × 10−20 kg ⋅ m/s.

(b) K = ( pc )2 + ( mc 2 ) 2 − mc 2 = 3.3 × 10−14 J = 0.21 MeV.
EVALUATE: (c) The result of part (b), about 2 × 105 eV, is many orders of magnitude larger than the
39.81.

potential energy of an electron in a hydrogen atom.
(a) IDENTIFY and SET UP: Δ xΔp x ≥ /2. Estimate Δ x as Δ x ≈ 5.0 × 10−15 m.
EXECUTE: Then the minimum allowed Δpx is Δpx ≈

2Δx

=

1.055 × 10−34 J ⋅ s
2(5.0 × 10−15 m)

= 1.1 × 10−20 kg ⋅ m/s.

(b) IDENTIFY and SET UP: Assume p ≈ 1.1 × 10−20 kg ⋅ m/s. Use Eq. (37.39) to calculate E, and then

K = E − mc 2 .

EXECUTE: E = (mc 2 ) 2 + ( pc) 2 . mc 2 = (9.109 × 10−31 kg)(2.998 × 108 m/s)2 = 8.187 × 10−14 J.

pc = (1.1 × 10−20 kg ⋅ m/s)(2.998 × 108 m/s) = 3.165 × 10−12 J.
E = (8.187 × 10−14 J)2 + (3.165 × 10−12 J)2 = 3.166 × 10−12 J.
K = E − mc 2 = 3.166 × 10−12 J − 8.187 × 10−14 J = 3.084 × 10−12 J × (1 eV/1.602 × 10−19 J) = 19 MeV.
(c) IDENTIFY and SET UP: The Coulomb potential energy for a pair of point charges is given by
Eq. (23.9). The proton has charge + e and the electron has charge – e.

ke2
(8.988 × 109 N ⋅ m 2 /C2 )(1.602 × 10−19 C) 2
=−
= −4.6 × 10−14 J = −0.29 MeV.
r
5.0 × 10−15 m
EVALUATE: The kinetic energy of the electron required by the uncertainty principle would be much larger
than the magnitude of the negative Coulomb potential energy. The total energy of the electron would be
large and positive and the electron could not be bound within the nucleus.
IDENTIFY: Apply the Heisenberg Uncertainty Principle. Let the uncertainty product have its minimum
possible value, so ΔxΔp x = /2.
SET UP: Take the direction of the electron beam to be the x -direction and the direction of motion
perpendicular to the beam to be the y -direction.
EXECUTE: U = −

39.82.

EXECUTE: (a) Δv y =

Δp y

2mΔy

=

1.055 × 10−34 J ⋅ s

= 0.12 m/s.
2(9.11 × 10−31 kg)(0.50 × 10−3 m)
(b) The uncertainty Δr in the position of the point where the electrons strike the screen is
Δp y x
x
Δr = Δv yt =
=
= 4.78 × 10−10 m.
m vx 2mΔy 2 K /m
EVALUATE: (c) This is far too small to affect the clarity of the picture.
m

=

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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.