A) \[1.1\times {{10}^{7}}\,V\]
B) \[2.1\times {{10}^{7}}\,V\]
C) \[3.1\times {{10}^{7}}\,V\]
D) \[0.15\times {{10}^{7}}\,V\]
Correct Answer: A
Solution :
[a] Charge on the tin nucleus = n.Q \[=(1.6\times {{10}^{-19}}\times 50)C=8\times {{10}^{-18}}C\] This charge is supposed to be concentrated at its centre. Hence, potential on the surface, \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}\cdot \frac{q}{r}\] \[=\frac{9\times {{10}^{9}}\times 8\times {{10}^{-18}}}{6.6\times {{10}^{-15}}}=1.1\times {{10}^{7}}\,\,V\]You need to login to perform this action.
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