Quiz4_solution_2D_2010

Quiz4_solution_2D_2010 - (1) (a) By Braggs Law, n =2 d sin...

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(1) (a) By Bragg’s Law, n λ =2 d sin θ (n=1,2,……) Lowest reflection corresponds to smallest θ , i.e. n=1 so λ = 2 (2.08. 10 -10 ) sin 15º m = 1.077. 10 -10 m E = hf = hc/ λ = (4.136.10 -15 eV.s) (3 . 10 8 m.s -1 )/ (1.077. 10 -10 m) = 11.52 . 10 3 eV = 11.52 KeV (b) By Compton’s formula λ ′ - λ = (h/m e c)(1 – cos ϕ ) = (6.626. 10 -34 J.s)/[ ( 9.1. 10 -31 kg)( 3.10 8 m/s) ] x (1 – cos 150º ) = 0.243. 10 -11 (1 + 0.866) m = 0.453. 10 -11 m Thus λ ′ = (1.077 + 0.0453) . 10 -10 m = 1.12. 10 -10 m so E = hc/ λ ′ = (4.136.10 -15 eV.s) (3 . 10 8 m.s -1 )/ (1.12. 10 -10 m) = 11.07 KeV (2) (a) n=9: r = 9 2 . a 0 = (81) (0.529. 10 -10 ) m = 4.28. 10 -9 m n=10: r = 10 2 . a 0 = (100) (0.529. 10 -10 ) m = 5.29. 10 -9 m (b) Classical Period of orbit T c = 2πr/v But v = Λ μρ = ν η so, T c = 2 πμρ 2 η
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Quiz4_solution_2D_2010 - (1) (a) By Braggs Law, n =2 d sin...

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