ch24_p117 - 117. (a) With V = 1000 V, we solve V= q 4o R...

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117. (a) With V = 1000 V, we solve V = q 4 πε o R 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 6.95 x 10 9 1 2 (3.7 x 10 8 ) = 38 seconds. (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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