Question

A copper slab of mass $2g$ contains $2\times {10}^{22}$ atoms. The charge on the nucleus of each atom is $29e$. What fraction of the electrons must be removed from the sphere to give it a charge of $+2\mu C$?

• A. $58\times {10}^{22}$
• B. $1.25\times {10}^{13}$
• C. $2.16\times {10}^{11}$
• D. $2.16\times {10}^{-11}$

Answer 1

The correct option is D $2.16\times {10}^{-11}$

Given a mass of copper slab $=2g$ contains $2\times {10}^{22}$ atoms.

Charge on nucleus of each atom $=29e$

We have to find the fraction of electrons that must be removed from sphere to give it $+2\mu C$ charge.

We solve it as follows:

Total number of electrons in sphere $=29\times 2\times {10}^{22}$

Number of electrons $=\frac{q}{e}$

$=\frac{2\times {10}^{-6}}{1.6\times {10}^{-19}}$

$=1.25\times {10}^{13}$

Fraction of electrons removed $=\frac{1.25\times {10}^{13}}{Totalnumberofelectronsinsphere}$

$=\frac{1.25\times {10}^{13}}{29\times 2\times {10}^{22}}$

$=2.16\times {10}^{-11}$