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hw24 - PHYS-1200 PHYSICS II HOMEWORK SOLUTIONS CLASS 24...

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PHYS-1200 PHYSICS II SPRING 2006 HOMEWORK SOLUTIONS CLASS 24. 39.06, 39.14, 39.16; Q41.03 39.06. 2 2 2 8 mL h n E n = . Then, 2 2 2 4 8 4 mL h E = , and 2 2 2 1 8 1 mL h E = . Since 1 4 E E E - = , we get, 2 2 2 2 2 2 2 2 2 2 2 2 2 2 8 15 8 ) 1 16 ( 8 ) 1 4 ( 8 1 8 4 mL h mL h mL h mL h mL h E = - = - = - = . 2 12 31 2 34 2 2 ) m 10 250 )( kg 10 11 . 9 ( 8 s) J 10 63 . 6 ( 15 8 15 - - - × × × = = mL h E E = 1.45 ×10 -17 J = 90.5 eV 39.14. a) The wave function for the ground state is: ψ π = 2 L L x sin . The probability that the particle is between x = 0 and x = L /4, is given by the integral: [ ] π π π π π π π π π π ψ 2 1 4 1 2 sin 2 1 4 1 0 4 2 sin 2 1 0 4 1 ) 4 / 0 ( 2 sin 2 1 1 2 cos 1 1 ) 4 / 0 ( 2 cos 1 2 1 2 sin 2 ) 4 / 0 ( 4 / 0 4 / 0 4 / 0 4 / 0 4 / 0 4 / 0 2 4 / 0 2 - = - = - - - = < < - = - = < < - = = = < < L L L x P L x L L x L dx L x L dx L L x P dx L x L dx L x L dx L x P L L L L L L L P (0 < x < L /4) = 0.091 b) Rather than repeat the integral for each area, it is instructive to look at a graph of ψ ² as a function of x . The graph below shows the three areas.
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It is apparent, from the symmetry, that the P (0 < x < L /4) = P (3 L /4 < x < L ). Since The probability of finding the particle somewhere in the well is equal to one, we must
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