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Q.
An $ \alpha - $ particle of energy 5 MeV is scattered through $ 180{}^\circ $ by a fixed uranium nucleus. The distance of the closest approach is of the order of
JamiaJamia 2007
Solution:
According to law of conservation of energy, kinetic energy of $ \alpha - $ particle = potential energy of $ \alpha - $ 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 $ $ 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 $