A nuclear physicist performed an experiment on the scattering of… | JEE Main 2022 (30 Jul, Shift 2)

Q9JEE Main 2022 · 30 JUL · SHIFT 2Medium

Atomic Models & Alpha-Particle Scattering

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 _______ $\times 10^{-16}\text{ m}$

(Take $\frac{1}{4\pi\epsilon_0} = 9 \times 10^{9}\ \frac{\text{N m}^2}{\text{C}^2}$ )

Numerical answer type. Work it out, then open the answer.

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Answer: 711

Nuclear radius $\approx$ distance of closest approach: $r_0 = \dfrac{k(2e)(Ze)}{K}$ , $K = 3.2\times10^6 e$ J.
$r_0 = \dfrac{9\times10^9 \times 158 \times 1.6\times10^{-19}}{3.2\times10^6}$
$r_0 = 7.11\times10^{-14}$ m $= 711\times10^{-16}$ m. Answer: 711.

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Atomic Models & Alpha-Particle Scattering question from JEE Main 2022 (30 Jul, Shift 2)

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