p39_040 - 40. (a) The fractional change is E E = = 1 (hc/)...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
40. (a) The fractional change is E E = ∆( hc/λ ) hc/λ = λ µ 1 λ = λ µ 1 λ 0 1 λ = λ λ 0 1= λ λ +∆ λ 1 = 1 λ/ λ +1 = 1 ( λ/λ C )(1 cos φ ) 1 +1 . If λ =3 . 0cm=3 . 0 × 10 10 pm and φ =90 , the result is E E = 1 (3 . 0 × 10 10 pm / 2 . 43 pm)(1 cos 90 ) 1 +1 = 8 . 1 × 10 11 . (b) Now λ = 500 nm = 5 . 00 × 10 5 pm and φ =90 ,so E E = 1 (5 . 00 × 10 5 pm / 2 . 43 pm)(1 cos 90 ) 1 +1 = 4 . 9 × 10 6 . (c) With λ =25pmand φ =90 , we Fnd E E = 1 (25 pm / 2 . 43 pm)(1 cos 90 ) 1 +1 = 8 . 9 × 10 2 . (d) In this case, λ = hc/E = 1240 nm · eV / 1 . 0MeV=1 . 24 × 10 3 nm = 1 . 24pm, so E E = 1 (1 . 24pm / 2 . 43 pm)(1
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

Ask a homework question - tutors are online