A) \[\text{9}\times \text{1}{{0}^{5}}\text{ V}\]
B) \[\text{8}\times \text{1}{{0}^{\text{6}}}\text{ V}\]
C) 80 V
D) 9 V
Correct Answer: B
Solution :
Electric potential at the surface of an atomic nucleus of radius r, is \[V=\frac{1}{4\,\,\pi \,{{\varepsilon }_{0}}}\frac{Ze}{r}\] Given, Z = 50, \[r=9\times {{10}^{-13}}cm=9\times {{10}^{-15}}m\], \[e=1.6\times {{10}^{-19}}C\] \[\therefore \] \[V=\frac{9\times {{10}^{9}}\times 50\times 1.6\times {{10}^{-19}}}{9\times {{10}^{-15}}}\] \[\Rightarrow \] \[V=8\times {{10}^{6}}\] volt
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