The radius of nucleus of silver (atomic number $=$ $47$) is $3.4 \times {10^{ - 14}}\,m$. The electric potential on the surface of nucleus is $(e = 1.6 \times {10^{ - 19}}\,C)$
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(a) $V = \frac{1}{{4\pi {\varepsilon _0}}}.\frac{{(Ze)}}{r} = 9 \times {10^9} \times \frac{{47 \times 1.6 \times {{10}^{ - 19}}}}{{3.4 \times {{10}^{ - 14}}}} = 1.99 \times {10^6}\,V$
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