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Question

Physics Question on Electrostatic potential

The radius of nucleus of silver (atomic number =47) is 3.4×10−14m.3.4\times {{10}^{-14}}m. The electric potential on the surface of nucleus is (e=1.6×10−19C)(e=1.6\times {{10}^{-19}}C)

A

1.99×106V1.99\times {{10}^{6}}V

B

2.9×106V2.9\times {{10}^{6}}V

C

4.99×106V4.99\times {{10}^{6}}V

D

0.99×106V0.99\times {{10}^{6}}V

Answer

1.99×106V1.99\times {{10}^{6}}V

Explanation

Solution

Charge on nucleus q=Ze=47×1.6×10−19q=Ze=47\times 1.6\times {{10}^{-19}} =7.52×10−18C=7.52\times {{10}^{-18}}C Potential at the surface of the nucleus V=14πε0.qrV=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{q}{r} =9×109×7.52×10−183.4×10−14=\frac{9\times {{10}^{9}}\times 7.52\times {{10}^{-18}}}{3.4\times {{10}^{-14}}} =1.99×106V=1.99\times {{10}^{6}}V