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

Answer:

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 = $\frac{k\times2e\times{Ze}}{r }$

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 = $\frac{9\times10^{9}\times{1.6}\times10^{-19}\times{1.6}\times10^{-19}\times2\times47}{22.9\times10^{-15} }$

K = 9.45x10⁻¹³ J