A) \[1.28\times {{10}^{-10}}m\]
B) \[1.92\times {{10}^{10}}m\]
C) \[2.56\times {{10}^{-10}}m\]
D) \[3.84\times {{10}^{10}}m\]
Correct Answer: C
Solution :
\[\frac{1}{2}m{{v}^{2}}+\frac{1}{2}m{{v}^{2}}=9\times {{10}^{9}}\times \frac{e\times e}{{{r}^{2}}}.\,\,r\] \[m{{v}^{2}}=\frac{9\times {{10}^{9}}\times {{(1.6\times {{10}^{-19}})}^{2}}}{r}\] \[r=\frac{9\times {{10}^{9}}{{(1.6\times {{10}^{-19}})}^{2}}}{m{{v}^{2}}}\] \[r=2.56\times {{10}^{-10}}m\]
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