410 7 Nuclear Physics – I
E 2 ′′=
2 E′′ 1
2 +α′′=
4 E 0
(2+α)(4)(1+α)(2+α)=
E 0
(1+α) 0
Photon energy after the third scattering through 60◦will beE 3 ′′=
2 E′′ 2
2 +α′′=
2 E 0
(1+α)(2+ 3 α)/(1+α)=
2 E 0
2 + 3 α
∴Total energy loss
ΔE′′=E 0 − 2 E 0 /(2+ 3 α)=(3αE 0 )/(2+ 3 α)(4)
Comparison of (2), (3) and (4) shows that the energy loss is equal for
Case (a) and (b), and each is greater than in case (c)7.59μc=σcρN 0 /A
μcx=σc(N 0 /A)(ρx)= 8. 1 × 10 −^24 × 6 × 1023 × 3. 7 / 27 = 6. 66
μph= 4 × 10 −^24 × 6 × 1023 × 3. 7 / 27 = 3. 288
μx=(μc+μph)χ= 6. 66 + 3. 288 = 9. 948
The intensity will be reduced by
I/I 0 =e−μx=e−^9.^948 = 4. 78 × 10 −^5
Ratio of intensities absorbed due to the Compton effect and due to the photo-
effect=(1−exp(−μcx))/(1−exp(−μphx))=(1−e−^6.^66 )/(1−e−^3.^288 )=
- 037
7.60 The change in wavelength in Compton scattering is given by
Δλ=mch(1−cosθ)
whereθ is the scattering angle.Δλwill be maximum forθ = 180 ◦in
which case
Δλ(max)= 2 h/mc= 4 c/mc^2
= 4 π× 197 .3 MeV-fermi/ 0 .511 MeV= 0. 0485 A ̊
7.61 The energyEof the Compton scattered photon by the incident photon of
energyE 0 is
E=E 0 /[(1+α(1−cosθ)]
whereα=E 0 /mc^2 andθis the scattering angle.
E/E 0 = 1 / 2 = 1 /[(1+α(1−cos 45◦)]
whenceα= 3. 415
OrE=(3.415((0.511)= 1 .745 MeV
7.62 Energy of the scattered gamma rays
E=E 0 /[1+α(1−cosθ)] (1)
Whereα=E 0 /mc^2
Ine+−e−annihilation, each gamma ray has energyE 0 = 0 .511 MeV
α= 0. 511 / 0. 511 = 1
Therefore (1) becomes
E= 0. 511 /(2−cosθ)
Emaxis obtained by puttingθ=0 andEminby puttingθ= 180 ◦. Thus
Emax= 0 .511 MeV
Emin= 0 .17 MeV