Phys 181b Problem Set 13 Solution

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

This preview shows pages 1–4. Sign up to view the full content.

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.

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

View Full Document
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 ∆= − = = = = ×
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.

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 07/18/2008 for the course PHYS 181 taught by Professor Stephenirons during the Spring '08 term at Yale.

### Page1 / 6

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online