Hello Student , this can be easily derived as shown below

Let m and v be the mass and velocity of alpha particle directed towards the centre of nucleus.

Kinetic Energy of alpha particle = K = 1/2 * mv²

Positive charge on the nucleus is Ze and charge of alpha particle is 2e

Electrostatic potential energy of the alpha particle when it is at a distance d from the centre of nucleus is U = 2Ze²/4πDε ₀

At closest approach Kinetic Energy of particle = Potential Energy

So >>> 1/2 * mv² = 2Ze²/4πDε ₀

D = 4Ze²/mv²4πε ₀

or D = Ze²/mv²πε ₀

Hope my answer was helpful. Do not memorize the formula remember the derivation using concept.