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:

पर्याय

• \[\frac{1}{\sqrt{m}}\]

• \[\frac{1}{m^2}\]

• m

• \[\frac{1}{m}\]

Answer 1

\[\frac{1}{m}\]

Explanation:

Let the distance of closest approach is r₀

\[\frac{1}{4\pi\varepsilon_{0}}\times\frac{(Ze)(2e)}{r_{0}}=\frac{1}{2}mv^{2}\Rightarrow r_{0}=\frac{1}{\pi\varepsilon_{0}}\times\frac{Ze^{2}}{mv^{2}}\]

\[r_0\propto\frac{1}{m}\]