ch24-p103 - 103. (a) With V = 1000 V, we solve V = q / 4 0...

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103. (a) With V = 1000 V, we solve 0 /4 , Vq R πε = where R = 0.010 m for the net charge on the sphere, and find q = 1.1 × 10 9 C. Dividing this by e yields 6.95 × 10 9 electrons that entered the copper sphere. Now, half of the 3.7 × 10 8 decays per second resulted in electrons entering the sphere, so the time required is 9 8 6.95 10 38 s (3.7 10 / s) / 2 × = × (b) We note that 100 keV is 1.6 × 10 14 J (per electron that entered the sphere). Using the given heat capacity, we note that a temperature increase of
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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