Nuclear Physics And Radioactivity Question 33
Question: $ \alpha - $ particles of energy 400 KeV are bombarded on nucleus of $ _{82}Pb $ . In scattering of $ \alpha - $ particles, its minimum distance from nucleus will be [RPET 1997]
Options:
A) 0.59 nm
B) 0.59 Å
C) 5.9 pm
D) 0.59 pm
Show Answer
Answer:
Correct Answer: D
Solution:
Suppose closest distance is r, according to conservation of energy. $ 400\times 10^{3}\times 1.6\times {{10}^{-19}}=9\times 10^{9}\frac{(ze),(2e)}{r} $
$ \Rightarrow 6.4\times {{10}^{-14}} $ $ =\frac{9\times 10^{9}\times (82\times 1.6\times {{10}^{-19}})\times (2\times 1.6\times {{10}^{-19}})}{r} $
$ \Rightarrow r=5.9\times {{10}^{-13}}m=0.59,pm $ .
BETA
