Question

An $\alpha $- particle moving with initial kinetic energy K towards a nucleus of atomic number z approaches a distance 'd' at which it reverses its direction. Obtain the expression for the distance of closest approach ’d’ in terms of the kinetic energy of $\alpha $- particle K.

Answer 1

Hint:We know that the nucleus consists of neutrons and protons with electrons revolving around it. So when an $\alpha $- particle approaches it, its kinetic energy gets converted to potential energy. This is to be kept in mind while solving the problem.

Formulas used:
Electric potential energy $U=\dfrac{k{{q}_{1}}{{q}_{2}}}{r}$, where q1, q2 are charged particles, r is the distance between them and k is the Coulomb’s constant whose value is $\dfrac{1}{4\pi {{\varepsilon }_{0}}}$

Complete step by step answer:
The electrical potential energy of a system of point charges can be interpreted as the work that is required to assemble this system of charges by bringing the particles together, as in the system from an infinite distance.
Here, the $\alpha $- particle approaches the nucleus and at a distance d, it gets deflected or the path is reversed. Hence, at this point, the kinetic energy of motion is converted into potential energy to bring the particle closer to the nucleus.
Kinetic Energy K is converted to PE at a distance d, that is at the distance of closest approach.
$U=\dfrac{k{{q}_{1}}{{q}_{2}}}{r}$
Here, ${{q}_{1}}=+2e$ since it is an alpha particle.
${{q}_{2}}=Ze$ (given)
And r=d.
Substituting values , we get
$\begin{align}
& K=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\times \dfrac{2Z{{e}^{2}}}{d} \\
& \therefore d=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\times \dfrac{2Z{{e}^{2}}}{K} \\
\end{align}$

Hence, we have obtained the value for the distance of closest approach.

Note:In this given condition, the distance of closest approach is also dependent on its impact parameter that is the perpendicular distance between the path of the alpha particle and the centre of the potential field.