Question

An Alpha Nucleus of energy $\overline 2$mv bombards a heavy nuclear target of charge$Ze$ . Then the distance of the closest approach for the alpha nucleus will be proportional to

Answer 1

The conservation of energy method for a system at the closest distance has to be used here. Thereafter, the formula for the initial and final potential energy of the system is to be included. With the initial and final kinetic energy of the system at the closest approach and then by equating it all together during the collision, the required relation for the closest distance of approach can be found.

Complete step-by-step solution:
Let the closest distance up to which the alpha particle approaches be $r$ .
At this distance, the kinetic energy of the alpha particle is zero.

Given: The Initial kinetic energy of the alpha particle, $K.{E_i}$=$\overline 2$​mv
Also, the initial potential energy of the system $P.{E_i}$ is assumed to be zero.
The atomic number of an alpha particle, ${Z_\alpha } = 2$
Mass of the system be $m$
By using the formula of Final potential energy of the system at the closest distance,
$P.{E_f}$​= $\dfrac{{K\left( {Ze} \right)2e}}{r}$where $K = \dfrac{1}{{4\pi {\varepsilon _0}}}$
By using the conservation of energy method we can get, the following relation
$K.{E_i} + P.{E_i} = K.{E_f} + P.{E_f}$
$\Rightarrow 2mv + 0 = 0 + \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{\left( {2e} \right)\left( {Ze} \right)}}{r}$
$\Rightarrow r = \dfrac{{K.4Z{e^2}}}{{2mv}}$(where$K = \dfrac{1}{{4\pi {\varepsilon _0}}}$​)
Hence $r \propto \dfrac{{K.2Z{e^2}}}{{mv}} \propto \dfrac{1}{m}$.

Note: The potential energy of a system of two atoms depends on the distance between them. At large distances, the energy is zero, which means there is no interaction. At distances of several atomic diameters, attractive forces dominate, whereas at very close approaches the force is repulsive, causing the energy to rise. When the $\alpha$particles reach the closest distance of the nucleus, they will come to rest and its kinetic energy will be completely transformed into potential energy that is given by the equation below $P.{E_f}$​= $\dfrac{{K\left( {Ze} \right)2e}}{r}$ where $K = \dfrac{1}{{4\pi {\varepsilon _0}}}$and $r$is the distance of the closest approach of the $\alpha$particles and e is the charge of the electron. Then by equating the conservation of energy method to both initial and final, we can arrive at the relation between the distance of the closest approach and the mass of the alpha nucleus.