Atoms And Nuclei Question 277

Question: An $ \alpha $ -particle of energy 5 MeV is scattered through $ 180{}^\circ $ by a fixed uranium nucleus. The distance of closest approach is of the order of

Options:

A) $ {{10}^{-12}}cm $

B) $ {{10}^{-10}}cm $

C) $ {{10}^{-14}}cm $

D) $ {{10}^{-15}}cm, $

Show Answer

Answer:

Correct Answer: A

Solution:

  • Distance of closest approach $ \text{Energy, }E=5\times 10^{6}\times 1.6\times {{10}^{-19}}J. $ $ r_{0}=\frac{Ze( 2e )}{4\pi {\varepsilon_{0}}( \frac{1}{2}mv^{2} )} $
    $ \therefore r_{0}=\frac{9\times 10^{9}\times ( 92\times 1.6\times {{10}^{-19}} )\times ( 2\times 1.6\times {{10}^{-19}} )}{5\times 10^{6}\times 1.6\times {{10}^{-19}}} $
    $ \Rightarrow r=5.2\times {{10}^{-14}}m=5.3\times {{10}^{-12}}cm. $


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