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: [ ] 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 - = - = - - - = < < - = -
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This note was uploaded on 04/07/2008 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.

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

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