A) \[1.99\times {{10}^{6}}V\]
B) \[2.9\times {{10}^{6}}V\]
C) \[4.99\times {{10}^{6}}V\]
D) \[0.99\times {{10}^{6}}V\]
Correct Answer: A
Solution :
Charge on nucleus \[q=Ze=47\times 1.6\times {{10}^{-19}}\] \[=7.52\times {{10}^{-18}}C\] Potential at the surface of the nucleus \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{q}{r}\] \[=\frac{9\times {{10}^{9}}\times 7.52\times {{10}^{-18}}}{3.4\times {{10}^{-14}}}\] \[=1.99\times {{10}^{6}}V\]
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