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Q.

A Copper atom has 29 electrons revolving around the nucleus. A copper ball contains $4 \times 10^{23}$ atoms. What fraction of the electrons be removed to give the ball a charge of $+ 9 .6 μC$ ?

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a

$~ 1.8 \times 10^{- 13}$

b

$~ 1.3 \times 10^{- 12}$

c

$~ 5.2 \times 10^{- 12}$

d

$6 \times 10^{- 10}$

answer is D.

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Detailed Solution

$Q = ne \Rightarrow 9.6 \times 10^{- 6} = n \times 1.6 \times 10^{- 19} \Rightarrow n = \frac{9.6 \times 10^{- 6}}{1.6 \times 10^{- 19}} = 6 \times 10^{13}$
$Fraction of e^{-} removed, \frac{n}{N_{0}} = \frac{6 \times 10^{13}}{4 \times 10^{23} \times 29} \approx 5.2 \times 10^{- 12}$

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