Question

During the Rutherford's $\alpha -$ particle scattering experiment:

• A. Distance of closest approach is of the order of ${10}^{-14}m$
• B. At the distance of closest approach, the kinetic energy of the $\alpha -$ particle is transformed into electrostatic potential
• C. The $\alpha$ particles moving towards the nucleus will stop moving and then start retracing its path
• D. Maximum number of $\alpha -$ particles went straight through the metal foil

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A Distance of closest approach is of the order of ${10}^{-14}m$
B At the distance of closest approach, the kinetic energy of the $\alpha -$ particle is transformed into electrostatic potential
C The $\alpha$ particles moving towards the nucleus will stop moving and then start retracing its path
D Maximum number of $\alpha -$ particles went straight through the metal foil

(a) In the Rutherford’s $alpha-$ particle scattering experiment, the distance of closest approach is ${10}^{-14}m$.
i.e.,$r_0=$ distance of closest approach$=\frac{1}{4\pi ϵ}\times \frac{2Z{e}^2}{E_k}$. . Further calculation give $r_0=$ ${10}^{-14}m$.
(b) At the distance of closest approach, the KE of $\alpha -$particle is equal to the electrostatic potential
i.e., $\frac{1}{2}m{v}^2=\frac{1}{4\pi ϵ}=\frac{Ze\times 2e}{r_0}$
(c) At the point of closest approach, since KE = electrostatic potential the $\alpha$ particle stops and then starts moving in backward direction.
(d) In Rutherford experiment, maximum number of particles did pass through the metal foil.