A copper atom has 29 electrons revolving around the nucleus. A copper ball contains $4 \times 10^{23}$ atoms. What fraction of the electron be removed to give the ball a charge of $+ 9.6 μC$ ?

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

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

$6 \times 10^{- 10}$

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

answer is D.

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Detailed Solution

Here , No of electrons = 29

Total no of atoms = 4 x 10 ²³

⇒ Total number of electrons ( N) = 29 x 4 x 10²³ = 116 x 10²³

Charge of the ball (q)= 9.6 μ C $\Rightarrow$ the number of electrons = 6x10¹³

fraction to be removed is $\frac{6 x 10^{13}}{116 x 10^{23}} ≃ 5.2 x 10^{- 12}$

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