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An $α$ -particle having energy 10MeV collides with a nucleus of atomic number 50. Then distance of closest approach will be

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detailed solution

Correct option is D

KE=10MeV=10×106ev; Z=50; q1=2e, q2=50eKE=14π∈0.q1q2d∴d=9×109×2e×50e10×106e =9×109×100×16×10−19107 =14.4×10−15m

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An α -particle having energy 10MeV collides with a nucleus of atomic number 50. Then distance of closest approach will be

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An $α$ -particle having energy 10MeV collides with a nucleus of atomic number 50. Then distance of closest approach will be