p40_049 - p ( a ) = 1 − e − 2 (1 + 2 + 2) = 1 − 5 e...

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49. According to Sample Problem 40-8, the probability the electron in the ground state of a hydrogen atom can be found inside a sphere of radius r is given by p ( r )=1 e 2 x ( 1+2 x +2 x 2 ) where x = r/a and a is the Bohr radius. We want r = a ,so x =1and
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Unformatted text preview: p ( a ) = 1 − e − 2 (1 + 2 + 2) = 1 − 5 e − 2 = 0 . 323 . The probability that the electron can be found outside this sphere is 1 − . 323 = 0 . 677. It can be found outside about 68% of the time....
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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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