Atomic StructureHard

Question

A proton and a deuteron are projected towards the stationary gold nucleus in di­fferent experiments with the same speed. The distance of closest approach will be

Options

A.same for both.
B.greater for proton.
C.greater for deuteron.
D.depends on speed.

Solution

Distance of closest approach, $r = \frac{K.q_{1}q_{2}}{\left( \frac{1}{2}mv^{2} \right)}$

From question: $r \propto \frac{1}{m}$

$\therefore\frac{r_{p}}{r_{d}} = \frac{m_{d}}{m_{p}} = \frac{2}{1}$

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