In the Rutherford scattering experiment, the distance of closest approach for an a-particle is do. If a-particle is replaced by a proton, then how much kinetic energy in comparison to a-particle will be required to have the same distance of Closest approach do ?

In the Rutherford scattering experiment, the distance of closest approach (d0d0) is determined by the balance between the electrostatic repulsion between the incident particle (alpha particle or proton) and the positively charged nucleus, and the kinetic energy of the incident particle.

Let's denote the kinetic energy required to achieve the distance of closest approach d0d0 for an alpha particle as KαKα, and the kinetic energy required for a proton to achieve the same distance of closest approach as KpKp.

The kinetic energy of a particle can be related to its charge and mass by the equation:

K=12mv2K=21mv2

Where:

• KK is the kinetic energy,
• mm is the mass of the particle,
• vv is the velocity of the particle.

In both cases, for the alpha particle and the proton, the electrostatic repulsion with the nucleus is the same, since both have the same charge (one charge unit). So, the kinetic energy required to achieve the same distance of closest approach is only dependent on the mass of the particle.

Given that the mass of an alpha particle (mαmα) is approximately 4 times the mass of a proton (mpmp), we can use the fact that kinetic energy is directly proportional to mass. Therefore, to achieve the same distance of closest approach d0d0, the kinetic energy required for the proton (KpKp) would be 1441 times the kinetic energy required for the alpha particle (KαKα).

In mathematical terms: Kp=14KαKp=41Kα

So, to achieve the same distance of closest approach, the kinetic energy required for the proton will be one-fourth of the kinetic energy required for the alpha particle.