47. The magnitude of the fractional energy change for the photon is given by
∆E
ph
E
ph
=
∆(hc/λ)
hc/λ
=
λ∆
1
λ
= λ
1
λ
−
1
λ + ∆λ
=
∆λ
λ + ∆λ
= β
where β = 0.10. Thus ∆λ = λβ/(1
− β). We substitute this expression for ∆λ in Eq. 39-11 and solve
for cos φ:
cos φ
=
1
−
mc
h
∆λ = 1
−
mcλβ
h(1
− β)
= 1
−
β(mc
2
)
(1
− β)E
ph
=
1
−
(0.10)(511 keV)
(1
− 0.10)(200 keV)
= 0.716 .
This leads to an angle of φ = 44
◦
.