Q. The radius of nucleus of silver (atomic number Z=47) is $3.4 \times 10^{- 14}m.$The electric potential on the surface of nucleus will be
A) $1.99 \times 10^{6}V$
B) $2.99 \times 10^{6}V$
C) $3.99 \times 10^{6}V$
D) None of these
Explanation
(a) Charge on nucleus
$\mathcal{Q =}Ze = 47 \times 1.6 \times 10^{- 19}$
$= 7.52 \times 10^{- 18}C$
Potential on the surface of nucleus is
$V = \frac{1}{4\pi\varepsilon_{0}}\frac{\mathcal{Q}}{r}$
$= \frac{9 \times 10^{9} \times 7.52 \times 10^{- 18}}{3.4 \times 10^{- 14}} = 1.99 \times 10^{6}V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.