Question

An alpha particle of energy 5MeV is scattered through 180 degree by a fixed uranium nucleus. The distance of the closest approach is of the order of
A. ${{10}^{-15}}cm$
B. ${{10}^{-13}}cm$
C. ${{10}^{-12}}cm$
D. ${{10}^{-19}}cm$

Answer 1

Hint: The key idea behind the closest approach of the $\alpha$(alpha) particle is that the total mechanical energy of the system is conserved. The total initial kinetic energy converts into final electrostatic potential energy when the $\alpha$particle momentarily stops due to electrostatic force of repulsion.

Formula Used:
Kinetic energy $KE=\dfrac{1}{2}m{{v}^{2}}$
Electrostatic Potential Energy $PE=\dfrac{{{q}_{1}}{{q}_{2}}}{4\pi {{\varepsilon }_{0}}r}$

Complete step by step answer:
Alpha particles are nuclei of helium atoms and, therefore, carry two units i.e. 2e of positive charge and have the mass of helium atom.
In our question $\alpha$particles of energy 5MeV are scattered through 180 degree by a fixed Uranium nucleus.
That means the entire Kinetic Energy of $\alpha$particle which is 5MeV is converted into electrostatic potential energy. Because the $\alpha$ particle momentarily stops due the effect of electrostatic repulsion when the $\alpha$particle just approaches a positively charged nucleus.

By energy conservation law
Kinetic Energy = Electrostatic Potential Energy
$5MeV=\dfrac{2Z{{e}^{2}}}{4\pi {{\varepsilon }_{0}}r}$
Where Z is the atomic number (or number of protons) of Uranium and Charge of $\alpha$particle is 2e.
$\begin{aligned} & r=\dfrac{2Z{{e}^{2}}}{4\pi {{\varepsilon }_{0}}KE} \\\ & r=\dfrac{2\times 92\times {{e}^{2}}\times 9\times {{10}^{9}}}{5\times {{10}^{6}}eV} \\\ & r=5.3\times {{10}^{-14}}m \\\ & r=5.3\times {{10}^{-12}}cm \\\ \end{aligned}$
This distance is called the distance of closest approach In our question the distance of closest approach is of the order ${{10}^{-12}}cm$.
Hence the option C is correct.

Note: In Rutherford $\alpha$ scattering experiment, the gold foil is placed instead of uranium. And the number of $\alpha$ particles rebound or deflect by more than 90 degree is only 1 in 8000. It suggests that the entire atom is empty and the positive protons and mass are concentrated in tiny nuclei.
While taking electrostatic potential energy due to $\alpha$particle and whatever may be the foil. Remember to take a number of protons (or atomic number) of foil. Which reside in the nucleus and also remember $\alpha$ has 2e.