The electric potential at the surface of an atomic nucleus $(Z = 50)$ of radius $9.0×{10^{ - 13}}\, cm$ is
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(b) $V = 9 \times {10^9} \times \frac{{50 \times 1.6 \times {{10}^{ - 19}}}}{{9 \times {{10}^{ - 15}}}} = 8 \times {10^6}\,V$
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