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The electric potential at the surface of an atomic nucleus (Z=50) of radius 9.0x10^-15 m

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9 x 10⁵ volt

9 volt

8 x 10⁶ volt

80 volt

answer is B.

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Detailed Solution

$V = \frac{1}{4 {πε}_{0}} \frac{q}{r}; V = \frac{9 \times 10^{9} \times (50 \times 1 .6 \times 10^{- 19})}{9 \times 10^{- 15}}$
=8 x 10⁶ volt

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