p40_028

# p40_028 - 28. We are looking for the values of the ratio...

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28. We are looking for the values of the ratio E n x ,n y ,n z h 2 / 8 mL 2 = L 2 Ã n 2 x L 2 x + n 2 y L 2 y + n 2 z L 2 z ! = ( n 2 x + n 2 y + n 2 z ) and the corresponding diFerences. (a) ±or n x = n y = n z =1 , the ratio becomes 1 + 1 + 1 = 3 . 00. (b) ±or n x = n y =2and n z =1 , the ratio becomes 4 + 4 + 1 = 9 . 00. One can check (by computing other ( n x ,n y ,n z ) values) that this is the third lowest energy in the system. One can also check that this same ratio is obtained for ( n x ,n y ,n z )=(2 , 1 , 2) and (1 , 2 , 2). (c) ±or n x = n y =1and n z =3 , the ratio becomes 1 + 1 + 9 = 11 . 00. One can check (by computing other ( n x ,n y ,n z ) values) that this is three “steps” up from the lowest energy in the system. One can also check that this same ratio is obtained for ( n x ,n y ,n z )=(1 , 3 , 1) and (3 , 1 , 1). If we take the diFerence between this and the result of part (b), we obtain 11 . 00 9 . 00 = 2 . 00. (d) ±or n x = n y =1and n z =2 , the ratio becomes 1 + 1 + 4 = 6 . 00. One can check (by computing
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