p39_083 - 6 . 63 10 34 J s s 2 (9 . 8 m / s 2 )(24 m) 1 2...

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83. The kinetic energy of the car of mass m moving at speed v is given by E = 1 2 mv 2 , while the potential barrier it has to tunnel through is U = mgh ,whe re h = 24 m. According to Eq. 39-21 and 39-22 the tunneling probability is given by T e 2 kL ,where k = r 8 π 2 m ( U E ) h 2 = s 8 π 2 m ( mgh 1 2 mv 2 ) h 2 = 2 π
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Unformatted text preview: 6 . 63 10 34 J s s 2 (9 . 8 m / s 2 )(24 m) 1 2 (20 m / s) 2 = 1 . 2 10 38 m 1 . Thus, 2 kL = 2(1 . 2 10 38 m 1 )(30 m) = 7 . 2 10 39 . One can see that T e 2 kL is essentially zero....
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