Question

A tin nucleus has charge $+50\, eV$. 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
$Q =50\, eV$
$=50 \times 1.6 \times 10^{-19}$
$=80 \times 10^{-19}$
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,
$=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{r}$
$=9 \times 10^{9} \times \frac{80 \times 10^{-19}}{10^{-12}}$
$=7.2 \times 10^{4} V$
(where the value of $\frac{1}{4 \pi \varepsilon_{0}}$ is $9 \times 10^{9} N - m ^{2} /{ }^{\circ} C$