Question

A copper atom has 29 electrons revolving around the nucleus. A copper ball contains 4x10^23 atoms. What fraction of electrons be removed to give the ball a charge of +9.6x10^-6 C?

Answer 1

According to the values

given in your question,

solution is as follows:

Charge on each atom=29 C
Total charge on the copper ball=29×4×10=1160 C
Thus,the number of electrons on the copper ball=1160(Since it is initially uncharged)
Number of electrons in 9.6 C
=9.6÷(1.6×10^−19)
=6×10^19

This is the number of electrons to be removed.

So, the fraction of electrons to be removed is given by
(6×10^19)÷1160
=(6000×10^16)÷(1160)
=5.17×10^16

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