Question

A tin nucleus has charge $+50eV$. If the proton is at $10^{-12}m$ from the nucleus. Then the potential at this position will be (charge on protons $=1.6\times 10^{-19}C$ ) :

• A. $7.2\times 10^{8}V$
• B. $3.6\times 10^{4}V$
• C. $14.4\times 10^{4}V$
• D. $7.2\times 10^{4}V$

Answer 1

Answer: (D)

Here : Charge on the nucleus
$Q=50eV$
$=50\times 1.6\times 10^{-19}$
$=80\times 10^{-19}$
Distance of proton from nucleus
$r=10^{-12}m$
Charge on proton $=1.6\times 10^{-19}C$
Potential of this position is given by,
$=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{Q}{r}$
$=9\times 10^9\times \frac{80\times 10^{-19}}{10^{-12}}$
$=7.2\times 10^{4}V$
(where the value of $\frac{1}{4\pi {\epsilon }_0}$ is $9\times 10^{9}N-m^2/{}^{\circ }C$