# Chapter-39 - 1523 Chapter 39 1. Since E n ∝ L – 2 in...

This preview shows pages 1–3. Sign up to view the full content.

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.

Unformatted text preview: 1523 Chapter 39 1. Since E n ∝ L – 2 in Eq. 39-4, we see that if L is doubled, then E 1 becomes (2.6 eV)(2) – 2 = 0.65 eV. 2. We first note that since h = 6.626 × 10 –34 J·s and c = 2.998 × 10 8 m/s, hc = × ⋅ × × = ⋅--- 6 626 10 2 998 10 1602 10 10 1240 34 8 19 9 . . . J s m / s J / eV m / nm eV nm. c hc h c hc h Using the mc 2 value for an electron from Table 37-3 (511 × 10 3 eV), Eq. 39-4 can be rewritten as E n h mL n hc mc L n = = 2 2 2 2 2 2 2 8 8 bg c h . The energy to be absorbed is therefore ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 2 2 2 2 2 4 1 2 2 2 2 3 4 1 15 15 1240eV nm 90.3eV. 8 8 8 511 10 eV 0.250nm e e h hc E E E m L m c L- ⋅ ∆ =- = = = = × 3. We can use the mc 2 value for an electron from Table 37-3 (511 × 10 3 eV) and hc = 1240 eV · nm by writing Eq. 39-4 as E n h mL n hc mc L n = = 2 2 2 2 2 2 2 8 8 bg c h . For n = 3, we set this expression equal to 4.7 eV and solve for L : L n hc mc E n = = ⋅ × = bg c h b g c hb g 8 3 1240 8 511 10 4 7 085 2 3 eV nm eV eV nm. . . 4. With m = m p = 1.67 × 10 – 27 kg, we obtain ( 29 ( 29 ( 29 2 34 2 2 2 21 1 2 2 27 12 6.63 10 J.s 1 3.29 10 J 0.0206eV. 8 8(1.67 10 kg) 100 10 m h E n mL---   ×     = = = × =       × ×   CHAPTER 39 1524 Alternatively, we can use the mc 2 value for a proton from Table 37-3 (938 × 10 6 eV) and hc = 1240 eV · nm by writing Eq. 39-4 as E n h mL n hc m c L n p = = 2 2 2 2 2 2 2 8 8 bg d i . This alternative approach is perhaps easier to plug into, but it is recommended that both approaches be tried to find which is most convenient. 5. To estimate the energy, we use Eq. 39-4, with n = 1, L equal to the atomic diameter, and m equal to the mass of an electron: ( 29 ( 29 ( 29( 29 2 2 34 2 2 10 2 2 31 14 1 6.63 10 J s 3.07 10 J=1920MeV 1.9 GeV. 8 8 9.11 10 kg 1.4 10 m h E n mL---- × ⋅ = = = × ≈ × × 6. (a) The ground-state energy is ( 29 ( 29 ( 29 2 34 2 2 2 18 1 2 2 31 12 6.63 10 J s 1 1.51 10 J 8 8(9.11 10 kg) 200 10 m 9.42eV. e h E n m L----   × ⋅     = = = ×     × ×     = (b) With m p = 1.67 × 10 – 27 kg, we obtain ( 29 ( 29 ( 29 2 34 2 2 2 22 1 2 2 27 12 3 6.63 10 J s 1 8.225 10 J 8 8(1.67 10 kg) 200 10 m 5.13 10 eV. p h E n m L-----   × ⋅     = = = ×       × ×     = × 7. According to Eq. 39-4 E n ∝ L – 2 . As a consequence, the new energy level E' n satisfies ′ = ′ F H G I K J = ′ F H G I K J =- E E L L L L n n 2 2 1 2 , which gives ′ = L L 2 . Thus, the ratio is / 2 1.41. L L ′ = = 8. Let the quantum numbers of the pair in question be n and n + 1, respectively. Then E n +1 – E n = E 1 ( n + 1) 2 – E 1 n 2 = (2 n + 1) E 1 ....
View Full Document

## This note was uploaded on 08/09/2010 for the course PHY 202 101A2 taught by Professor Prof.yang during the Summer '10 term at National Taiwan University.

### Page1 / 26

Chapter-39 - 1523 Chapter 39 1. Since E n ∝ L – 2 in...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online