1000 Solved Problems in Modern Physics

408 7 Nuclear Physics – I

Fig. 7.15

hν 0 /c=hνcosθ/c+Pecosφ (2)

0 =−hνsinθ/c+Pesinφ (3)

wherePeis the electron momentum

Re-arranging (2) and (3) and squaring

Pe^2 cos^2 φ=

(

hν 0

c

−

hνcosθ

c

) 2

(4)

Pe^2 sin^2 φ=

(

h

c

νsinθ

) 2

(5)

Add (4) and (5) and using the relativistic equation

c^2 Pe^2 =T^2 + 2 Tmc^2 =h^2 (ν 02 +ν^2 − 2 ν 0 νcosθ)(6)

Eliminating T between (1) and (2) and simplifying we get

E=E 0 /[1+α(1−cosθ)] (7)

(b)T = E 0 −E = E 0 −E 0 /[1+α(1−cosθ)]= [αE 0 (1−cosθ)]/

[1+α(1−cosθ)] (8)

(c) From (2) and (3), we get

Cotφ=(ν 0 −νcosθ)/νsinθ

With the aid of (7), and re-arranging we find tan(θ/2) = (1+α)

tanφ (9)

7.54 Fractional shift in frequency is

Δν/ν 0 =

1

1 + Mc

2

2 hν 0 sin^2 (θ 2 )

hν 0 =Eγ= 1 , 241 /λnm= 1 , 241 / 0. 1 = 12 ,410 eV= 0 .01241 MeV

Putθ= 180 ◦andMC^2 = 938 .3 MeV to obtain

(Δν/ν 0 )max=

1

1 + 2 ×^9380. 01241.^3

= 2. 645 × 10 −^5

7.55 The energy of scattered photon will be

hν=hν 0 /[1+α(1−cosθ)]