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Q.

A Copper atom has 29 electrons revolving around the nucleus. A copper ball contains 4×1023 atoms. What fraction of the electrons be removed to give the ball a charge of +9.6 μC?

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a

~ 1.8×10−13

b

~ 1.3×10−12

c

~ 5.2×10−12

d

6×10−10

answer is D.

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Detailed Solution

Q=ne ⇒9.6×10−6=n×1.6×10−19 ⇒n=9.6×10−61.6×10−19=6×1013
 Fraction of e−removed, nN0=6×10134×1023×29≈5.2×10−12

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