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Q. An $\alpha$ - particle accelerated through $V$ volt is fired towards a nucleus. Its distance of closest approach is $r$. If a proton is accelerated through the same potential and fired towards the same nucleus, the distance of closest approach of proton will be:

AMUAMU 2018Atoms

Solution:

K.E = P.E $\Rightarrow $ $q_1 V = \frac{1}{4 \pi \varepsilon_0} \frac{q_1q_2}{r}$
Independent of charge of the particle projected