HW1Ans - Incomplete

HW1Ans Incomplete - HaoLi Sept.2010 2 < x >= x x t dx = 2 2 0 2 2 x 2 x 2 xL 4 x L2 4 x L x sin 2)dx = sin cos = 2 L L L 4 8 L 32 L02 2 2 < x >= x

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Radiation Physics Homework 1 Solutions Hao Li Sept. 2010 2. 2 22 2 2 2 2 0 0 2 4 4 ( , ) sin ( ) [ sin( ) cos( )] 48 3 2 2 x xx L x L x L x t d d x LL L L L ππ π <> = Ψ = = = ∫∫ 2 2 2 3 2 2 2 2 23 0 0 2 2 4 8 4 ( , ) s i n () [ c o s s i n ] 61 6 6 4 38 x L x x L L x x t d x x d x L L L = Ψ = = =− 3. If we assume the box is an infinite square well, its kinetic energy is 2 2 ,, 2 88 8 nml ee e hh h En m l mL =+ + When n,m,l = 1, its minimum kinetic energy is: 34 2 21 7 1,1,1 31 9 2 (6.626 10 ) 3 1 1.81 10 113 89 . 1 11 0 ( 0 . 0 ) EJ J e V −− × = × = ⋅× × 4. a). 2 2 2 000 s in 1 rr ddd r θϕθ Ψ= ∫∫∫ 2 22 2 sin 1 r Ne r d d d r α = 2 2 00 0 sin 1 r Nr e d r d d ϕθ θ = 2 3 4 1 4 N = 3 N = b).
Background image of page 1

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

View Full DocumentRight Arrow Icon
33 2 2 22 3 2 2 000 0 0 0 ( sin cos ) sin sin cos rr x er r d d d r r e d r d d ππ αα θ ϕθ ϕ θϕ −− <> = = ∫∫∫ Because 2 0 cos 0 d π ϕϕ = , 0 x = . 2 2 3 2 0 0 0 3 4 sin sin ( )(2 )(2) 82 r e r r d d dr r e dr d d θϕθ θθ α ∞∞ = ⋅⋅ = == 2 2 2 2 4 2 0 0 0 3 52 sin sin ( )(2 )(2) 4 r e r r d d dr r e dr d d = = c).
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 10/30/2010 for the course MP 200 taught by Professor Guna during the Fall '10 term at Duke.

Page1 / 3

HW1Ans Incomplete - HaoLi Sept.2010 2 < x >= x x t dx = 2 2 0 2 2 x 2 x 2 xL 4 x L2 4 x L x sin 2)dx = sin cos = 2 L L L 4 8 L 32 L02 2 2 < x >= x

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online