p40_013

# p40_013 - 10 9 nm 2(1 0 eV(1240 eV ย nm 2 โ 1 2 โ 12 ร...

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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 ×
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Unformatted text preview: 10 9 nm) 2 (1 . 0 eV) (1240 eV ยท nm) 2 โ 1 2 โ 12 ร 10 18 . (c) At this value of n , the energy is E n = 1240 2 8(511 ร 10 3 )(3 . ร 10 9 ) 2 ( 6 ร 10 18 ) 2 โ 6 ร 10 18 eV . (d) Since E n mc 2 = 6 ร 10 18 eV 511 ร 10 3 eV ร 1 , the energy is indeed in the relativistic range....
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## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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