A) 129
B) 130
C) 139
D) 120
Correct Answer: C
Solution :
Given, \[E=10\text{ }MeV=10\times 1.6\times {{10}^{-13}}J\] \[=1.6\times {{10}^{-12}}J\] \[{{r}_{0}}=4\times {{10}^{-14}}m\] Using, \[{{r}_{0}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{2Z{{e}^{2}}}{E}\] \[Z=\frac{({{r}_{0}}E)4\pi {{\varepsilon }_{0}}}{2({{e}^{2}})}\] \[=\frac{(4\times {{10}^{-14}})(1.6\times {{10}^{-12}})}{2\times {{(1.6\times {{10}^{-19}})}^{2}}}\times \frac{1}{9\times {{10}^{9}}}\] \[=139\]
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