The Wave Nature of Particles
3911
(b)
34
7
27
35
distance
0.80 m
7.3 10 s(1 y/3.156 10 s)
2.3 10 y
velocity
1.1 10
m/s
t
−
==
=
×
×
=
×
×
Since you walk through doorways much more quickly than this, you will not experience diffraction effects.
EVALUATE:
A 1 kg object moving at 1 m/s has a de Broglie wavelength
34
6.6 10
m,
λ
−
=×
which is exceedingly
small. An object like you has a very, very small
at ordinary speeds and does not exhibit wavelike properties.
39.56.
(a)
19
7
2.58 eV
4.13 10
J, with a wavelength of
4.82 10
m
482 nm
hc
E
E
−−
×
=
=
×
=
(b)
34
28
9
7
(6.63 10
J s)
6.43 10
J
4.02 10
eV.
22
(
1
.
6
4
1
0
s
)
h
E
π
t
π
−
−
×⋅
Δ=
=
=
×
=
×
Δ×
(c)
, so (
)
0, and
Eh
c
E
E
EE
λλ
=Δ
+
Δ
=
Δ
=
Δ
, so
28
71
6
7
19
6.43 10
J
(
4
.
8
21
0 m
)
7
.
5
01
0 m 7
.
5
0 nm
.
4.13 10
J
−
−
−
⎛⎞
×
Δ= Δ
=
×
=
×
=
×
⎜⎟
×
⎝⎠
39.57.
IDENTIFY:
The electrons behave as waves whose wavelength is equal to the de Broglie wavelength.
SET UP:
The de Broglie wavelength is
=
h/mv
, and the energy of a photon is
E = hf = hc/
.
EXECUTE: (a)
Use the de Broglie wavelength to find the speed of the electron.
()
(
)
34
31
9
6.626 10
J s
9.11 10
kg 1.00 10 m
h
v
m
−
××
= 7.27
×
10
5
m/s
which is much less than the speed of light, so it is nonrelativistic.
(b)
Energy conservation gives
eV =
½
mv
2
.
V = mv
2
/2
e
= (9.11
×
10
–31
kg)(7.27
×
10
5
m/s)
2
/[2(1.60
×
10
–19
C)] = 1.51 V
(c)
K = eV
=
e
(1.51 V) = 1.51 eV, which is about ¼ the potential energy of the NaCl crystal, so the electron would
not be too damaging.
(d)
E = hc/
=
(4.136
×
10
–15
eV s)(3.00
×
10
8
m/s)/(1.00
×
10
–9
m) = 1240 eV
which would certainly destroy the molecules under study.
EVALUATE:
As we have seen in Problems 39.10 and 39.43, when a particle and a photon have the same
wavelength, the photon has much more energy.
39.58.
sin
sin , and
(
)
(
2
), and so
θθ
hp
h
mE
′
′′
′
′
=
arcsin
sin
2
h
mE
⎛
⎞
′ =
⎜
⎟
′
⎝
⎠
.
34
11
31
3
19
(6.63 10
J s)sin 35.8
arcsin
20.9
(3.00 10
m) 2(9.11 10
kg)(4.50 10 )(1.60 10
J eV)
θ
−
+
−
°
′
°
×
×
39.59.
(a)
The maxima occur when
2s
i
n
d
θ
m
=
as described in Section 38.7.
(b)
2
hh
p
mE
.
34
10
37
19
(6.63 10
J s)
1.46 10
m
0.146 nm
2(9.11 10
kg)(71.0 eV) 1.60 10
J/eV
−
−
×
=
.
1
sin
(Note: This
2
m
θ
m
d
−
=
is the order of the maximum, not the mass.)
10
1
11
(1)(1.46 10
m)
sin
53.3 .
2(9.10 10
m)
−
−
−
×
⇒
=°
×
(c)
The work function of the metal acts like an attractive potential increasing the kinetic energy of incoming
electrons by
.
e
φ
An increase in kinetic energy is an increase in momentum that leads to a smaller wavelength. A
smaller wavelength gives a smaller angle
(see part (b)).
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '06
 Buchler
 Physics, Diffraction, Uncertainty Principle, kx − ωt

Click to edit the document details