Problem 3.2
Find the de Broglie wavelength of (a) an electron whose speed is 1x10
8
m/s and (b) an electron whose speed is
2x10
8
m/s.
γ
v
( )
1
1
v
2
c
2
−
:=
v
part1
1.0 10
8
⋅
:=
v
part2
2.0 10
8
⋅
:=
Part a.
Do we need to worry about relativity?
Let's do this non-relativistically and relativistically, and see if there
is a significant difference
Nonrelativistic:
Relativistic
λ
non
m v
,
(
)
h
m v
⋅
:=
λ
rel
m v
,
(
)
h
γ
v
( ) m
⋅
v
⋅
:=
λ
non
m
electron
v
part1
,
(
)
7.278
10
12
−
×
=
λ
rel
m
electron
v
part1
,
(
)
6.861
10
12
−
×
=
wavelength units are meters
Percent error P:
P
λ
non
m
electron
v
part1
,
(
)
λ
rel
m
electron
v
part1
,
(
)
−
λ
rel
m
electron
v
part1
,
(
)
100
⋅
:=
P
6.066
=
Because a 6% error is introduced by the nonrelativistic calculation, we need to use
the relativistic result.
Useful constants, etc.:
h
6.63 10
34
−
⋅
:=
m
electron
9.11 10
31
−
⋅
:=
c
3 10
8
⋅
:=
=====================================================================================
Physics 107
Problem 3.1
Compare the linear momenta, total energy, and kinetic energy of a photon and a particle that have the same
wavelength.
Because
λ
h
p
:=
regardless of whether the object is a photon
or a particle, if the wavelengths are the same, then the momenta p are the same.
Problem 3.11 shows that the particle's total energy is nearly the same as the energy of a photon of the
same wavelength PROVIDED THE PARTICLE'S TOTAL ENERGY GREATLY EXCEEDS ITS REST
ENERGY; i.e. when the particle's kinetic energy is very large; it is moving very fast.
From chapter 1,
E
m
0
2
c
4
⋅
p
2
c
2
⋅
+
:=
In this problem, both the photon and the particle have the same momemtum p, so p
2
c
2
is the same for
both.
Because some of the particle's total energy is tied up in its rest energy m
0
2
c
4
, and none of the
photon's total energy is tied up in rest energy, the particle must have less kinetic energy than the
photon has total energy.
=====================================================================================
Physics 107
1

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E
m
electron
c
2
⋅
1
v
2
c
2
−
:=
We can use equation 1.22 to calculate the velocity
joules
E
8.816
10
14
−
×
=
convert to joules
E
E 1.60
⋅
10
19
−
⋅
:=
eV
E
5.51
10
5
×
=
E
40 10
3
⋅
0.511 10
6
⋅
+
:=
The electron's total energy is its 40-keV kinetic energy plus its 0.511 MeV rest energy.
In units of eV,
we need to use relativistic corrections for this magnitude of speed
v
cl
1.185
10
8
×
=
v
cl
2 KE
⋅
m
electron
:=
if this gives a v near c, then we need to go back and calculate a
relativistic velocity
KE
1
2
m
⋅
v
2
⋅
:=
KE
40 10
3
⋅
1.60
⋅
10
19
−
⋅
:=
Part b.
Because the velocity is greater than in part a, we clearly need a relativistic calculation.
λ
rel
m
electron
v
part2
,
(
)
2.712
10
12
−
×
=
=====================================================================================
Physics 107
Problem 3.3
Find the de Broglie wavelength of a 1 mg grain of sand blown by the wind at 20 m/s.
This is an extremely lightweight piece of matter.
Maybe its wavelength will be observable??
λ
sand
m v
,
(
)
h
m v
⋅
:=
λ
sand
1 10
6
−
⋅
20
,
(
)
3.315
10
29
−
×
=
(using the fact that one milligram is 10
-3
grams, or 10
-6
kg)
This wavelength is in meters.
This is many orders of magnitude smaller than a typical size of a
nucleus.
We'll never "see" that a grain of sand has wave properties.

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