Nuclear Physics And Radioactivity Question 24
Question: An a-particle of 5 MeV energy strikes with a nucleus of uranium at stationary at an scattering angle of 180o. The nearest distance upto which a-particle reaches the nucleus will be of the order of [IIT 1981; AIEEE 2004]
Options:
A) $ 1,\overset{o}{\mathop{A}}, $
B) $ {{10}^{-10}}cm $
C) $ {{10}^{-12}}cm $
D) $ {{10}^{-15}}cm $
Show Answer
Answer:
Correct Answer: C
Solution:
At closest distance of approach Kinetic energy = Potential energy
$ \Rightarrow 5\times 10^{6}\times 1.6\times {{10}^{-19}}=\frac{1}{4\pi {\varepsilon_{0}}}\times \frac{(ze)(2e)}{r} $ For uranium z= 92, so $ r=5.3\times {{10}^{-12}}cm $
BETA
