Quiz4_solution_2D_2010

# Quiz4_solution_2D_2010 - (1(a By Braggs Law n =2 d...

This preview shows pages 1–2. Sign up to view the full content.

(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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 2

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online