Question

The electric potential at the surface of an atomic nucleus $(Z=50)$ of radius $9.0 \times 10^{-13} cm$ is

• A. $9 \times 10^{5} V$
• B. $9 \times 10^{6} V$
• C. $80\, V$
• D. $9\, V$

Answer 1

Answer: (B)

$9 \times 10^{6} V$

Answer 2

Electric potential at the surface of an atomic nucleus of radius $r$, is
$V=\frac{1}{4 \pi \varepsilon_{0}} \frac{Z e}{r}$
Given, $Z=50$,
$r=9 \times 10^{-13} cm =9 \times 10^{-15} m$,
$e=1.6 \times 10^{-19} C$
$\therefore V=\frac{9 \times 10^{9} \times 50 \times 1.6 \times 10^{-19}}{9 \times 10^{-15}}$
$\Rightarrow V=8 \times 10^{6}$ volt