Environmental Engineering Reference
In-Depth Information
this type of calculation, the key is to realise that we can make very good use of
Eq. (7.36) to make progress, i.e. we know that
E
e
−
p
e
c
2
m
e
c
4
.
=
(7.41)
If we can evaluate the left hand side of this expression in terms of photon variables
only then we will have succeeded in our task. This we can do since Eq. (7.38) and
Eq. (7.39) together imply that
c
2
p
e
E
cos
θ)
2
E
2
sin
2
θ
=
−
+
(E
(7.42)
(using cos
2
φ
sin
2
φ
+
=
1) and Eq. (7.40) implies that
E
e
E
+
m
e
c
2
)
2
.
=
(E
−
(7.43)
Subtracting these last two equations and using Eq. (7.41) gives
E
+
m
e
c
2
)
2
E
cos
θ)
2
E
2
sin
2
θ
m
e
c
4
,
(E
−
−
(E
−
−
=
which can be re-arranged to give
2
EE
(
1
2
m
e
c
2
(E
E
).
−
cos
θ)
=
−
This equation can be easily solved for
E
but we prefer to re-write it as
1
E
=
1
E
+
(
1
−
cos
θ)
m
e
c
2
.
(7.44)
In actual fact Compton measured the wavelength of the incoming and outgoing
light and used the de Broglie relationship to relate the energy of a photon to its
wavelength, i.e.
E
=
hc/λ
,where
h
is Planck's constant. Consequently, Eq. (7.44)
becomes
hc(
1
cos
θ)
m
e
c
2
−
λ
=
λ
+
.
(7.45)
This is our final answer: Special Relativity and Quantum Mechanics together lead
to a very definite prediction for the shift in wavelength of the scattered light as a
function of the angle
θ
and this prediction is strikingly confirmed by Compton's
original data, which we show in Figure 7.6. We plot the shift in wavelength
λ
=
λ
−
−
cos
θ
. The data are to be compared to the prediction of
Eq. (7.45) which is also shown as the straight line.
λ
as a function of 1
PROBLEMS 7
10
−
27
7.1 A proton has mass equal to 1
.
673
×
kg. Use this to determine the mass
of the proton in units of MeV/
c
2
.
