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

◦

.