JEE Main 2020 — Atomic Physics Question with Solution

Rutherford carried out an experiment in which he bombarded a thin sheet of gold foil with $\alpha$-particles and then analysed the trajectory of these particles after they collided with the gold foil.

For a detector at a specific angle with respect to the incident beam, the number of particles per unit area striking the detector is given by the Rutherford formula,

$Y\left(\theta \right)=\frac{N_{i}nL{Z}^2{k}^2{e}^4}{4r^2{\left(KE\right)}^2{\sin }^4\left({\displaystyle \frac{\theta }{2}}\right)}$

Here, $N_i=$ Number of incident alpha particles,

$n=$ Atoms per unit volume in target,

$L=$ Thickness of target,

$Z=$ Atomic number of target,

$e=$ Electron charge,

$k=$ Coulomb's constant,

$r=$ Distance between target and detector,

$KE=$ Kinetic energy of alpha particles,

$\theta =$ Scattering angle.

Therefore, $Y\propto \frac{1}{{\sin }^4\left({\displaystyle \frac{\theta }{2}}\right)}$