CBSE Class 12 · 2017 · Delhi · Modern PhysicsShort Answer

Define the distance of closest approach. An

α

-particle of kinetic energy

K

is bombarded on a thin gold foil. The distance of closest approach is

r

. What will be the distance of closest approach for an

α

-particle of double the kinetic energy?

Answer: $r_{0} \propto 1/ K$ , so distance becomes $r /2$ .

Solution

Distance of closest approach: the least distance from the centre of the nucleus to which a head-on

α

-particle approaches, where its entire kinetic energy has been converted into electrostatic potential energy (its velocity is momentarily zero).

K = \frac{1}{4 π ε _{0}} \frac{2 Z e ^{2}}{r _{0}} \Rightarrow r_{0} \propto \frac{1}{K} .

Doubling the kinetic energy halves the distance of closest approach:

r_{0}^{'} = \frac{r}{2}

CBSE Class 12 Board · 2017 · Delhi · Subjective