In a Rutherford scattering experiment, an alpha particle is accelerated toward a stationary silver nucleus. If the closest approach distance is 22.9 fm, what is the initial kinetic energy (in J) of the alpha particle?

Initial Kinetic energy of alpha particle is  9.45x10⁻¹³ J .

Explanation:

The distance at which the initial kinetic energy of the particle is equal to the potential energy is known as closest distance. As it is Rutherford scattering, so it is a coulomb potential energy.

Let K be the initial kinetic energy of alpha particle and r be the closest approach distance. So,

Initial Kinetic Energy = Coulomb Potential Energy

K =

Here, k is constant, e is charge of electron and Z is the atomic number of silver.

Put 9x10⁹ N m²/C² for k, 1.6x10⁻¹⁹ C for e, 47 for Z and 22.9x10⁻¹⁵ m for r in the above equation.

K =

K = 9.45x10⁻¹³ J

double the distance = divide strength by 4