A) \[1\,\overset{o}{\mathop{A}}\,\]
B) \[{{10}^{-10}}cm\]
C) \[{{10}^{-12}}cm\]
D) \[{{10}^{-15}}cm\]
Correct Answer: C
Solution :
At closest distance of approach Kinetic energy = Potential energy \[\Rightarrow 5\times {{10}^{6}}\times 1.6\times {{10}^{-19}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{(ze)(2e)}{r}\] For uranium z= 92, so \[r=5.3\times {{10}^{-12}}cm\]
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