A) \[1\overset{\text{o}}{\mathop{\text{A}}}\,\]
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 \[\alpha \]-particle at distance of closest approach ie, \[\frac{1}{2}m{{v}^{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{r}\] \[\therefore \] \[5MeV=\frac{9\times {{10}^{9}}\times (2e)\times (92e)}{r}\] \[\left( \because \,\,\frac{1}{2}m{{v}^{2}}=5MeV \right)\] \[\Rightarrow \] \[r=\frac{9\times {{10}^{9}}\times 2\times 92\times {{(1.6\times {{10}^{-19}})}^{2}}}{5\times {{10}^{6}}\times 1.6\times {{10}^{-19}}}\] \[\therefore \] \[r=5.3\times {{10}^{-14}}m\approx {{10}^{-12}}cm\]
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