A copper ball contains 4x10²³atoms. What fraction of the electrons be removed to give the ball a charge of 9.6 micro coloumbs?

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Solution

According to the values given in your question, solution is as follows:
Charge on each copper atom=29 C
Total charge on the
copper ball
=29×4×10^23
=11610^23 C
Thus,the number of electrons on the copper ball
=11610^23(Since it is initially uncharged)
9.6 micro coloumbs=
9.610^-6
Number of electrons in
9.610^-6 coloumbs
=Q/e
=9.610^-6/1.610^-19
=610^13
This is the number of electrons to be removed
So the fraction of electrons to be removed is given by
610^13/11610^23

=.05110^-10
=51*10^-13

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A copper ball contains 4x1023 atoms. What fraction of the electrons be removed to give the ball a charge of 9.6 micro coloumbs?

A copper ball contains 4x10²³atoms. What fraction of the electrons be removed to give the ball a charge of 9.6 micro coloumbs?