This preview shows pages 1–3. Sign up to view the full content.
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 nonrelativistically and relativistically, and see if there
is a significant difference
Nonrelativistic:
Relativistic
λ
non
mv
,
h
⋅
:=
λ
rel
,
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.:
h6
.
6
3
1
0
34
−
⋅
:=
m
electron
9.11 10
31
−
⋅
:=
c3
1
0
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,
Em
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
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentE
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
−
⋅
:=
E5
.
5
11
0
5
×
=
E4
0
1
0
3
⋅
0.511 10
6
⋅
+
:=
The electron's total energy is its 40keV 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
2KE
⋅
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
mv
,
h
⋅
:=
λ
sand
110
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.
This is the end of the preview. Sign up
to
access the rest of the document.
 Fall '09
 Abra
 Electron, pH

Click to edit the document details