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← JEE Main 2022 · 30 Jul · Shift 2
PhysicsInteger+4 / -1Question 21 of 90

JEE Main 2022 · 30 Jul · Shift 2, Physics question 21

A nuclear physicist performed an experiment on the scattering of alpha particles by thin metal foil of gold. The alpha particles were accelerated up to kinetic energy 3.2 Mev. The estimation of the radius of the nucleus of gold element (Z = 79) by him is _______ ×10−16\times10^{-16} m

(Take 14πϵo=9×109\dfrac1{4\pi\epsilon_o}=9\times10^9 N m2^2/C2^2)

Answer and worked solution

Official answer: 711

Closest approach. r=k(2e)(79e)K=9×109×158×(1.6×10−19)23.2×106×1.6×10−19=711×10−16r=\dfrac{k(2e)(79e)}K=\dfrac{9\times10^9\times158\times(1.6\times10^{-19})^2}{3.2\times10^6\times1.6\times10^{-19}}=711\times10^{-16} m.

Answer: 711

Topics:Rutherford model
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