Physics Circle home Student Login

Home›Chapters›Atomic Structure PYQs›Atomic Models & Alpha-Particle Scattering›Question 9

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

Atomic Models & Alpha-Particle Scattering question from 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=79Z = 79) by him is _______ ×10−16 m\times 10^{-16}\text{ m}

(Take 14πϵ0=9×109 N m2C2\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.

Show answer and solution

Answer: 711

  1. Nuclear radius ≈\approx distance of closest approach: r0=k(2e)(Ze)Kr_0 = \dfrac{k(2e)(Ze)}{K}, K=3.2×106eK = 3.2\times10^6 e J.
  2. r0=9×109×158×1.6×10−193.2×106r_0 = \dfrac{9\times10^9 \times 158 \times 1.6\times10^{-19}}{3.2\times10^6}
  3. r0=7.11×10−14r_0 = 7.11\times10^{-14} m =711×10−16= 711\times10^{-16} m. Answer: 711.
Report a mistake

More questions on Atomic Models & Alpha-Particle Scattering