1. Calculate the de Broglie wavelength for an electron with a kinetic energy of 20 eV and 200
keV.
Answer:
The 20 eV electron is nonrelativistic, so we have
E
k
=
p
2
/
2
m
and
p
=
√
2
mE
k
giving a wavelength of
λ
=
h
p
=
h
√
2
mE
k
=
hc
√
2
mc
2
E
k
=
1240
eV
·
nm
r
2
·
(511
×
10
3
eV
)
·
(20
eV
)
= 0
.
274
nm
.
The 200 keV electron energy is close enough to the rest mass that we must use a relativistic
approach. We know
E
k
= (
γ
−
1)
mc
2
so
γ
= 1 +
E
k
mc
2
= 1 +
200
keV
511
keV
= 1
.
39
and then from
γ
≡
1
/
√
1
−
β
2
we can find
β
=
r
1
−
1
/γ
2
=
r
1
−
1
/
(1
.
39
2
) = 0
.
695
We can then use
λ
=
h/p
and
p
=
γmv
=
γβmc
to find
λ
=
h
γβmc
=
hc
γβmc
2
=
1240
eV
·
nm
1
.
39
·
0
.
695
·
(511
×
10
3
eV
)
= 0
.
0025
nm
.
2. Consider an atom with a nucleus consisting of 2 protons, and instead of an electron a muon
(207 times as heavy as an electron, or
m
μ
= 207
m
e
). Calculate the energy of the ground
state and the first excited state, and the wavelength of light associated with the transition.
Answer:
The reduced mass is
m
r
=
2
m
p
·
m
μ
2
m
p
+
m
μ
=
2
m
p
·
207
m
e
2
m
p
+ 207
m
e
=
2
·
(939
×
10
3
keV
)
·
207
·
(511
keV
)
2
·
(939
×
10
3
keV
) + 207
·
(511
keV
)
= 1
.
00
×
10
5
keV
.
Since
E
0
=
me
4
/
8
ǫ
2
0
h
2
scales with
m
, we can scale
E
0
with
m
e
to the reduced mass by
multiplying by
m
r
/m
e
. The Bohr energy then becomes
E
n
=
−
Z
2
n
2
E
0
m
r
m
e
=
−
1
n
2
2
2
(13
.
60
eV
)
1
.
00
×
10
5
keV
511
keV
=
−
1
n
2
1
.
065
×
10
4
eV
so that the energy of the
n
= 1
state is 10,650 eV while the energy of the
n
= 2
state is 2,660
eV. The energy difference is 7,990 eV giving a photon wavelength of
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.
 Fall '01
 Rijssenbeek
 Physics, Energy, Kinetic Energy, ev, Schr¨ dinger equation

Click to edit the document details