Question

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?

Answer 1

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

Step-by-step 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 =
`\frac{k*2e*{Ze}}{r }`

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 =
`\frac{9*10^(9)*{1.6}*10^(-19)*{1.6}*10^(-19)*2*47}{22.9*10^(-15) }`

K = 9.45x10⁻¹³ J