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  4. α-particles are projected towards the nuclei of the different metals, with the same kinetic energy. The distance of closest approach is minimum for

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Q. $\alpha$-particles are projected towards the nuclei of the different metals, with the same kinetic energy. The distance of closest approach is minimum for

Atoms

Solution:

$r= \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{(KE)} \, \Rightarrow \, r \alpha q_2$