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  4. The electric potential at the surface of an atomic nucleus (Z=50) of radius 9.0 × 10-15 m is:

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Q. The electric potential at the surface of an atomic nucleus $(Z=50)$ of radius $9.0 \times 10^{-15} m$ is:

JIPMERJIPMER 2003Electric Charges and Fields

Solution:

$V=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{R} =\frac{1}{4 \pi \varepsilon_{0}} \frac{(Z e)}{R} $
$=9 \times 10^{9} \times \frac{50 \times 1.6 \times 10^{-19}}{9.0 \times 10^{-15}} $
$=8 \times 10^{6} V $