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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