A) \[7.2\times {{10}^{8}}V\]
B) \[3.6\times {{10}^{4}}V\]
C) \[14.4\times {{10}^{4}}V\]
D) \[7.2\times {{10}^{4}}V\]
Correct Answer: D
Solution :
Here: Charge on the nucleus \[Q=50eV\] \[=50\times 1.6\times {{10}^{-19}}\] \[=80\times {{10}^{-19}}\] Distance of proton from nucleus \[r={{10}^{-12}}m\] Charge on proton \[=1.6\times {{10}^{-19}}C\] Potential of this position is given by, \[=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{Q}{r}\] \[=9\times {{10}^{9}}\times \frac{80\times {{10}^{-19}}}{{{10}^{-12}}}\] \[=7.2\times {{10}^{4}}V\] (where the value of\[\frac{1}{4\pi {{\varepsilon }_{0}}}\]is\[9\times {{10}^{9}}N-{{m}^{2}}/{}^\circ C\]
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