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7 Nuclear Physics – I
Fig. 7.15
h ν 0 /c = h ν cos θ/c + P e cos ϕ
(2)
0 = −h ν sin θ/c + P e sin ϕ
(3)
where P e is the electron momentum
Re-arranging (2) and (3) and squaring
P e
2 cos
2
ϕ =
hν 0
c
−
hν cos θ
c
2
(4)
P e
2 sin
2
ϕ =
h
c
ν sin θ
2
(5)
Add (4) and (5) and using the relativistic equation
c
2 P e
2
= T
2
+ 2T mc
2
= h
2 (ν 0
2
+ ν
2
− 2ν 0 ν cos θ )
( 6 )
Eliminating T between (1) and (2) and simplifying we get
E = E 0 /[1 + α(1 − cos θ)]
(7)
(b) T = E 0 − E = E 0 − E 0 /[1 + α(1 − cos θ )] = [α E 0 (1 − cos θ )]/
[1 + α(1 − cos θ)]
(8)
(c) From (2) and (3), we get
Cot ϕ = (ν 0 − ν cos θ)/ν sin θ
With the aid of (7), and re-arranging we find tan(θ/2) = (1 + α)
tan ϕ
(9)
7.54 Fractional shift in frequency is
Δν/ν 0 =
1
1 +
Mc 2
2hν 0 sin
2 (
θ
2 )
hν 0 = E γ = 1, 241/λ nm = 1, 241/0.1 = 12, 410 eV = 0.01241 MeV
Put θ = 180
◦ and MC
2
= 938.3 MeV to obtain
(Δν/ν 0 ) max =
1
1 +
938.3
2×0.01241
= 2.645 × 10
−5
7.55 The energy of scattered photon will be
h ν = h ν 0 /[1 + α(1 − cos θ)]