Electrostatics

JEE Main level, numerical answer, medium

175 of 311

Question 175 of 311

The electric potential at the surface of an atomic nucleus (Z=50)(Z=50) of radius 9×10−13cm9\times {10}^{-13}cm is α×106V\alpha \times {10}^{6}V. What is the value of α\alpha?
(Charge of proton 1.6×10−19C1.6\times {10}^{-19}C)

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Answer: 8

The electric potential at the surface of an atomic nucleus can be written as
V=kQR=k⋅(Ze)R...(1)V=\dfrac{kQ}{R} =\dfrac{k\cdot \left(Ze\right)}{R}...\left(1\right)
From equation (1), it follows that
V=9×109×50×1.6×10−199×10−13×10−2V=8×106V\begin{matrix}V & = & \dfrac{9\times {10}^{9}\times 50\times 1.6\times {10}^{-19}}{9\times {10}^{-13}\times {10}^{-2}}V \\ & = & 8\times {10}^{6}V\end{matrix}
Hence, α=8\alpha =8.

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