Question 1

When an \alpha-particle of mass m moving with velocity v bombards on a heavy nucleus of charge Ze, its distance of closest approach from the nucleus depends on m as

Accepted Answer

\frac{1}{m}. At the closest approach, all kinetic energy converts to potential energy:

\frac{1}{2}mv^2 = \frac{2Ze^2}{4\pi\varepsilon_0 r_0}

r_0 \propto \frac{1}{m}

Question 2

In a Rutherford scattering experiment when a projectile of charge Z_1 and mass M_1 approaches a target nucleus of charge Z_2 and mass M_2, the distance of closest approach is r_0. The energy of the projectile is

Accepted Answer

directly proportional to Z_1 Z_2. At the distance of closest approach the kinetic energy is completely converted to potential energy:

\frac{1}{2}M_1u^2 = \frac{1}{4\pi\varepsilon_0}\frac{Z_1Z_2}{r_0}

So the energy of the projectile is directly proportional to Z_1Z_2.

Question 3

In Rutherford scattering experiment, what will be the correct angle for \alpha-scattering for an impact parameter, b = 0?

Accepted Answer

180°. The impact parameter is related to the scattering angle by b \propto \cot\frac{	heta}{2}

For b = 0: \cot\frac{	heta}{2} = 0 \Rightarrow \frac{	heta}{2} = 90° \Rightarrow 	heta = 180°

(head-on collision, the particle retraces its path).

Alpha-Particle Scattering and Rutherford Nuclear Model of Atom — NEET Physics PYQ MCQs with Solutions

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