ch38-p035 - 35. (a) From Eq. 38-11 = h (1 cos ) = (2.43...

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

View Full Document Right Arrow Icon
(c) The change in energy for a photon with λ = 590 nm is given by ph 2 15 8 2 6 (4.14 10 eV s)(2.998 10 m/s)(2.43pm) (590nm) 8.67 10 eV . hc hc E ∆λ ⎛⎞ ∆= ⎜⎟ λλ ⎝⎠ ×⋅× =− × (d) For an x-ray photon of energy E ph = 50 keV, ∆λ remains the same (2.43 pm), since it is independent of E ph . (e) The fractional change in wavelength is now 3 2 15 8 ph (50 10 eV)(2.43pm) 9.78 10 . / (4.14 10 eV s)(2.998 10 m/s) hc E ∆λ ∆λ × == = × λ× × (f) The change in photon energy is now ∆λ ∆λ ∆λ Eh c hc E ph ph = + F H G I K J F H G I K J + + F H G I K J 11 1 λ λ α where = ∆λ / λ . With E ph = 50 keV and = 9.78 × 10 –2 , we obtain E ph = –4.45 keV. (Note that in this case 0.1 is not close enough to zero so the approximation
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 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

Ask a homework question - tutors are online