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Unformatted text preview: Publisher's Solution for HW # 8 CONCEPTUAL QUESTIONS 25.1. 1 decreases. As the crystal is compressed, the spacing d between the planes of atoms decreases. The Bragg condition is 2 cos m m d = so as d decreases, cos m must increase. But cos increases as decreases. 25.2.(a) a b c E E E because the energy per photon depends only on the frequency so / . E hf hc = = The smaller wavelengths correspond to higher frequencies. (b) c b a N N N because the powers are equal, there must be more photons when the energy per photon is less. 25.3. 2 2 1 1 1 1 / 1 / 2 2 E hc E hc = = = 25.5. Fast electrons will have a shorter wavelength leading to less diffraction spreading and better resolution. 25.7. Because 2 2 2 8 n h E n mL = we see that for a given n , n E is inversely proportional to 2 L . If L is doubled then n E is decreased by a factor of 4. So the new 19 1 1 10 J. E = 25.8. It is the same, or 20 1.0 10 J. 1 1 1 2 2 H He H 2 2 8 8(4 ) 2 h h E E E m L L m = = = EXERCISES AND PROBLEMS 25.4.Model: The angles of incidence for which diffraction from parallel planes occurs satisfy the Bragg condition. Solve: The Bragg condition is 2 cos , m d m = where m = 1, 2, 3, For first and second order diffraction, ( 29 1 2 cos 1 d = , ( 29 2 2 cos 2 d = Dividing these two equations, ( 29 ( 29 1 1 2 2 1 1 cos 2 cos 2cos cos 2cos68 41 cos  = = = = 25.7.Model: The angles corresponding to the various diffraction orders satisfy the Bragg condition. Solve: The Bragg condition is 2 cos m d m = , where m = 1, 2, 3, The maximum possible value of m is the number of possible diffraction orders. The maximum value of cos m is 1. Thus, ( 29 ( 29 2 0.180 nm 2 2 4.2 0.085 nm d d m m = = = = We can observe up to the fourth diffraction order....
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This note was uploaded on 12/26/2011 for the course PHYSICS 270 taught by Professor Drake during the Fall '08 term at Maryland.
 Fall '08
 drake

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