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. $
BETA
