A) \[2.3\,\,\times \,\,1{{0}^{-10}}m\]
B) \[2.3\,\,\times \,\,1{{0}^{-14}}m\]
C) \[4.8\,\,\times \,\,1{{0}^{-10}}m\]
D) \[4.8\,\,\times \,\,1{{0}^{-14}}m\]
Correct Answer: B
Solution :
As, \[E=0.5\,Me\,V=0.5\times 1.6\times 1{{0}^{-}}^{13}\,J\] Now, \[E=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{(Ze)(e)}{{{r}_{0}}}\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,{{r}_{0}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Z{{e}^{2}}}{E}=\frac{9\times {{10}^{9}}\times 79{{(1.6\times {{10}^{-19}})}^{2}}}{5\times 1.6\times {{10}^{-13}}}\]\[=\,\,2.3 \times \,\,1{{0}^{-}}^{14}\,m\]
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