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Question
When an α-particle of mass m moving with velocity ν bombards on a heavy nucleus of charge Ze, its distance of closest approach from the nucleus depends on m as:
Options
\[\frac{1}{\sqrt{m}}\]
\[\frac{1}{m^2}\]
m
\[\frac{1}{m}\]
MCQ
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Solution
\[\frac{1}{m}\]
Explanation:
Let the distance of closest approach is r0
\[\frac{1}{4\pi\varepsilon_{0}}\times\frac{(Ze)(2e)}{r_{0}}=\frac{1}{2}mv^{2}\Rightarrow r_{0}=\frac{1}{\pi\varepsilon_{0}}\times\frac{Ze^{2}}{mv^{2}}\]
\[r_0\propto\frac{1}{m}\]
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