Atomic StructureHard
Question
Two particles A and B having same e/m ratio are projected towards silver nucleus in different experiments with the same speed. The distance of closest approach will be
Options
A.same for both.
B.greater for A.
C.greater for B.
D.depends on speed.
Solution
$r = \frac{K.q_{1}q_{2}}{\left( \frac{1}{2}mv^{2} \right)}$
As $\left( \frac{q_{1}}{m} \right)$, q2 and v, all are same, r is same.
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