Phys 181b Problem Set 13 Solution

Phys 181b Problem Set 13 Solution - 4. With mp = 1.67 10 27...

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4. With m p = 1.67 × 10 – 27 kg, we obtain () 2 34 2 2 22 1 1 2 2 12 6.63 10 J.s 1 3.29 10 J 0.0206eV. 8 8 100 10 m p h En mL m §· × ¨¸ == = × = ©¹ × Alternatively, we can use the mc 2 value for a proton from Table 37-3 (938 × 10 6 eV) and the hc = 1240 eV · nm value developed in problem 83 of Chapter 38 by writing Eq. 39-4 as E nh mL c mc L n p 2 2 2 8 8 b g di . This alternative approach is perhaps easier to plug into, but it is recommended that both approaches be tried to find which is most convenient.
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6. We can use the mc 2 value for an electron from Table 37-3 (511 × 10 3 eV) and the hc = 1240 eV · nm value developed in problem 83 of Chapter 38 by writing Eq. 39-4 as E nh mL c mc L n == 22 2 2 2 8 8 b g c h . The energy to be absorbed is therefore () 2 41 2 2 3 15 15 1240eV nm 90.3eV. 8 8 8 511 10 eV 0.250nm e e h hc EE E mL mc L ∆= − = = = = ×
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14. We follow Sample Problem 39-3 in the presentation of this solution. The integration result quoted below is discussed in a little more detail in that Sample Problem. We note that the arguments of the sine functions used below are in radians.
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This note was uploaded on 07/18/2008 for the course PHYS 181 taught by Professor Stephenirons during the Spring '08 term at Yale.

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Phys 181b Problem Set 13 Solution - 4. With mp = 1.67 10 27...

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