p40_013 - 10 9 nm) 2 (1 . 0 eV) (1240 eV nm) 2 1 2 12 10 18...

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

View Full Document Right Arrow Icon
13. We can use the mc 2 value for an electron from Table 38-3 (511 × 10 3 eV) and the hc value developed in problem 3 of Chapter 39 by writing Eq. 40-4 as E n = n 2 h 2 8 mL 2 = n 2 ( hc ) 2 8( mc 2 ) L 2 . (a) With L =3 . 0 × 10 9 nm , the energy diFerence is E 2 E 1 = 1240 2 8(511 × 10 3 )(3 . 0 × 10 9 ) 2 ( 2 2 1 2 ) =1 . 3 × 10 19 eV . (b) Since ( n +1) 2 n 2 =2 n +1,wehave E = E n +1 E n = h 2 8 mL 2 (2 n +1)= ( hc ) 2 8( mc 2 ) L 2 (2 n +1) . Setting this equal to 1 . 0 eV, we solve for n : n = 4( mc 2 ) L 2 E ( hc ) 2 1 2 = 4(511 × 10 3 eV)(3 . 0 ×
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 10 9 nm) 2 (1 . 0 eV) (1240 eV nm) 2 1 2 12 10 18 . (c) At this value of n , the energy is E n = 1240 2 8(511 10 3 )(3 . 10 9 ) 2 ( 6 10 18 ) 2 6 10 18 eV . (d) Since E n mc 2 = 6 10 18 eV 511 10 3 eV 1 , the energy is indeed in the relativistic range....
View Full Document

Ask a homework question - tutors are online