A) \[1{AA}\]
B) \[{{10}^{-10}}cm\]
C) \[{{10}^{-12}}cm\]
D) \[{{10}^{-15}}cm\]
Correct Answer: C
Solution :
According to law of conservation of energy, kinetic energy of\[\alpha -\]particle = potential energy of a-particle at distance of closest approach i.e., \[\frac{1}{2}m{{v}^{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{r}\] \[\therefore \,\,\,5\,\,MeV\,=\frac{9\times {{10}^{9}}\,\times \,(2e)\,\times (92\,e)}{r}\] \[\left( \because \frac{1}{2}m{{v}^{2}}=5\,MeV \right)\] \[\Rightarrow \]\[r=\frac{9\times {{10}^{9}}+2\times 92\times {{(1.6\times {{10}^{-19}})}^{2}}}{5\times {{10}^{6}}\times 1.6\times {{10}^{-19}}}\] \[\therefore \]\[r=5.3\times {{10}^{-14}}m\approx {{10}^{-12}}cm\]
You need to login to perform this action.
You will be redirected in
3 sec
