Question

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

• A. $6.16\times {10}^{-11}$
• B. $2.16\times {10}^{-11}$
• C. $9.16\times {10}^{-11}$
• D. $2.16\times {10}^{-21}$

Answer 1

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

Given,

mass of Sphere is 2 gm

Number of atoms is $2\times {10}^{22}$

Charge Given to sphere is $2\mu C$

We Know that

$Q=ne$

$Numberofelectron\left(n\right)=\frac{Q}{e}$

$n=\frac{2\mu C}{1.6\times {10}^{-19}}$ [Because Charge of electron is $1.6\times {10}^{-19}$]

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

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

Now Total number of electron in sphere is

$29\times 2\times {10}^{22}$

$=58\times {10}^{22}$

Therefore,

Fraction of electron removed can be calculated as,

$\frac{1.25\times {10}^{13}}{Totalnumberofelectroninsphere}$

$=\frac{1.25\times {10}^{13}}{58\times {10}^{22}}$

$=2.155\times {10}^{-11}\approx 2.16\times {10}^{-11}$

Hence,Option (B) is correct option.