Homework 1
1
11/1/2005
Homework I and Solutions
Problems:
I.1
Show that
1.
e
2
/(4
πε
0
) = 1.44 MeV fm.
2.
α
=
1/137
Solution:
1.
e
2
/(4
πε
0
) = (1.602E–19 C)/(4*3.142*8.85E–12 F/m) e = 1.440E–9
eV m = 1.44 MeV fm
2.
α≡
e
2
/(4
πε
0
) / (
ħ
c
) = 1.440 MeV fm / 197.3 MeV fm = 1/137.0
I.2
Show that the process
e
−
→
e
−
+
γ
is forbidden (in free space) by energy and momentum conser
vation. Similarly, show that the process
γ
→
e
−
+
e
+
is forbidden in free space.
Solution:
Energymomentum conservation would be violated in the first process for a photon with nonzero
energy
E
and momentum
E/c
. This is immediately obvious when one considers the process in the
rest frame of the initial electron; any nonzero energy photon would involve a nonzero momen
tum, which would lead to a net energymomentum increase in the final state. The process only
occurs for energetic electrons in the Coulomb field of an nucleus, which takes up some recoil
momentum in the process.
The second process is forbidden because of energymomentum conservation as well. Assume the
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 Fall '05
 MichaelRijssenbeek
 Physics, Particle Physics, Work, Quark, Meson, Quark model, MeV fm

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