Question

A tin nucleus has charge $+50e$. If the proton is at $10^{-12} m$ from the nucleus, then the potential at this position, is (charge on proton is $1 \times 6 \times 10^{-19} C)$

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

Answer 1

Answer: (D)

Given: Charge $(Q) = 50e = 50 \times \left(1 .6×10 ^{-19} \right) C$;
Distance of proton from nucleus $\left(r\right) = 10^{-12}\,m$ and charge on proton $\left(e\right) = 1 .6 \times 10^{-19} C$. We know that potential
$\left(V\right)=\frac{1}{4\pi\,\varepsilon_{0}}\times\left(\frac{Q}{r}\right)=9\times10^{9}\times\frac{50\times1×6\times10^{-19}}{10^{-12}}$
$=7.2 \times 10^{4}\,V.$