When an alpha-particle is acceleration by a PD of 3 volt, its energy is

Suppose an alpha particle accelerated by a potential of V volt is allowed to collide with a nucleus whose atomic number is Z, then the distance of closest approach of alpha particle to the nucleus is

A protons , a deuteron and an alpha-particle are accelerated through the same p.d of V volt. The velocities acquired by them are in the ratio

An alpha -particle is accelerated through a.p.d of 10^(6) volt the K.E. of particle will be

An alpha -particle is accelerated through a.p.d of 10^(6) volt the K.E. of particle will be

One can define an alpha-Volt ( alpha V ) to be the energy acquired by an a particle when it is accelerated by a potential of 1 Volt. For this problem you may take a proton to be 2000 times heavier than an electron. Then