hw24 - PHYS-1200 PHYSICS II SPRING 2005 HOMEWORK SOLUTIONS...

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Unformatted text preview: PHYS-1200 PHYSICS II SPRING 2005 HOMEWORK SOLUTIONS CLASS 24. 39.06, 39.13, 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 39.13. The probability of finding the electron between x 1 and x 2 is equal to: = 2 1 2 ) ( x x dx x P . According to the hint, this can be replaced by: P = ( x ) 2 x , or = = L x L x x L x L P 2 2 sin 2 sin 2 . a) = = = 4 sin 10 ....
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This note was uploaded on 09/17/2008 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 SPRING 2005 HOMEWORK SOLUTIONS...

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