# Distance of Closest Approach

An  α-particle which moves straight towards the nucleus in head on direction reaches the nucleus i.e., it moves close to a distance r0 .

As the  α-particle approaches the nucleus, the electrostatic repulsive force due to the nucleus increase and kinetic energy of the alpha particle goes on converting into the electrostatic potential energy. When whole of the kinetic energy is converted into electrostatic potential energy, the α-particle cannot further move towards the nucleus but returns back on its initial path i.e α-particle is scattered through an angle of 180°. The distance of α-particle from the nucleus in this stage is called as the distance of closest approach and is represented by r0.

Let m and v be the mass and velocity of the α-particle directed towards the centre of the nucleus. Then kinetic energy of the α-particle is

$\displaystyle K = \frac{1}{2} m v^2$

Because the positive charge on the nucleus is Ze and that on the α-particle 2e , hence the electrostatic potential energy of the α-particle, when at a distance r0 from the centre of the nucleus, is given by

$\displaystyle U = \frac{1}{4 \pi \epsilon_0} \frac{(Z e ) (2 e)}{r_0}$

Because at distance of Closest approach ; kinetic energy of the α-particle appears as its potential energy, hence, K = U

$\displaystyle \frac{1}{2} m v^2 = \frac{1}{4 \pi \epsilon_0} \frac{(Z e ) (2 e)}{r_0}$

$\displaystyle r_0 = \frac{1}{4 \pi \epsilon_0} \frac{(Z e ) (2 e)}{\frac{1}{2} m v^2}$

### Impact Parameter (b)

The perpendicular distance of the initial velocity vector of the α-particle from centre of the nucleus is called ‘ Impact parameter ‘

$\displaystyle b = \frac{Z e^2 cot(\frac{\theta}{2})}{4 \pi \epsilon_0 (\frac{1}{2} m v^2) }$