1000 Solved Problems in Modern Physics

384 7 Nuclear Physics – I 7.17 In the elastic scattering of deuterons of 11.8 MeV from 82 Pb^208 , the differen- tial cross-sect ...

7.2 Problems 385 7.26 The following counting rates (in arbitrary units) were obtained whenαparti- cles were scattered through 18 ...

386 7 Nuclear Physics – I 7.36 If the mean range of 10 MeV protons in lead is 0.316 mm, calculate the mean range of 20 MeV deute ...

7.2 Problems 387 7.47 The range of 5 MeVα’s in air at NTP is 3.8 cm. Calculate the range of 10 MeV α’s using the Geiger–Nuttal l ...

388 7 Nuclear Physics – I 7.57 X-rays are Compton scattered at an angle of 60◦. If the wavelength of the scattered radiation is ...

7.2 Problems 389 7.67 Estimate the thickness of lead (density 11.3gcm−^3 ) required to absorb 90% of gamma rays of energy 1 MeV. ...

390 7 Nuclear Physics – I 7.2.6 Pair Production .................................... 7.75 Calculate the maximum wavelength ofγ-r ...

7.2 Problems 391 7.83 Calculate the spread in energy of the 661 keV internal conversion line of^137 Cs due to the thermal motion ...

392 7 Nuclear Physics – I 7.89 Determine the amount of^21084 Po necessary to provide a source ofα-particles of 10 millicuries st ...

7.2 Problems 393 7.98 Natural Uranium, as found on earth, consists of two isotopes in the ratio of^23592 U/^23892 U= 0 .7%. Assu ...

394 7 Nuclear Physics – I Why is the transition (^17) F→ (^17) O+e−+ν(5/ 2 +→ 5 / 2 +) called a super-allowed transition? 7.107 ...

7.3 Solutions 395 (becausemandMare oppositely directed in the CMS, and the recoil angle of the proton in the CMS is always twice ...

396 7 Nuclear Physics – I As triton will be emitted in the opposite direction in the CMS θt∗= 111 ◦ Using formula (1) again, wit ...

7.3 Solutions 397 7.7 Let the alpha particle of massmmoving with velocityv 0 and momentum p 0 collide with proton of massm. Afte ...

398 7 Nuclear Physics – I dΩ∗/dEp= 4 π/E 0 (2) And dσp/dΩ∗=σ/ 4 π (3) as the scattering is isotropic in the CMS. Using (2) and ( ...

7.3 Solutions 399 σ= ∫ σ(θ)dΩ= ∫π 0 (R^2 /4). 2 πsinθdθ=πR^2 This is called geometric cross-section. 7.3.2 RutherfordScattering. ...

400 7 Nuclear Physics – I Therefore, Mean free pathλ= 1 /Σ= 0 .2cm Fraction of scattered particles=Probability of scattering thr ...

7.3 Solutions 401 Fig. 7.14Collision of a heavy charged particle with an electron From the velocity triangle which is isosceles ...

402 7 Nuclear Physics – I NowR 0 = 1. 44 zZ/T= 1. 44 × 2 × 13 / 4. 5 = 8. 32 Hencermin= 1. 5 × 8. 32 = 12 .48 fm 7.25 The minimu ...

7.3 Solutions 403 7.30 The fraction of particles scattered in brass at angles exceedingθis given byΔN/N =(π/4)(1. 442 /T^2 ) ( 0 ...