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 $+ 9 .6 μC$ ?

see full answer

Want to Fund your own JEE / NEET / Foundation preparation ??

Take the SCORE scholarship exam from home and compete for scholarships worth ₹1 crore!*

An Intiative by Sri Chaitanya

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

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

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

$6 \times 10^{- 10}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

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