Question

An $\alpha$ particle of $10$ MeV is moving forward for a head-on collision. What will be the distance of the closest approach from the nucleus of the atomic number $Z=50$? Choose the correct option from the given options.
$\begin{aligned} & A)14.4\times {{10}^{-16}}m \\\ & B)1.7\times {{10}^{-7}}m \\\ & C)1.5\times {{10}^{-12}}m \\\ & D)14.4\times {{10}^{-15}}m \\\ \end{aligned}$

Answer 1

In this case the kinetic energy of $\alpha$ particle decreases and its potential energy increases continuously as it approaches the nucleus. At the distance of the closest approach the kinetic energy totally gets converted into the potential energy. We need to equate this and then we need to determine the distance of the closest approach.

Formula used: ${{r}_{0}}=\dfrac{2Z{{e}^{2}}}{4\pi {{\varepsilon }_{0}}K}$

Complete step-by-step solution
Now as the $\alpha$ particle comes closer to the nucleus, it has to do work against the positive coulomb force of the nucleus and its potential energy continuously increases at the expense of its kinetic energy. Let the distance of closest approach is ${{r}_{0}}$. if $Z$ is the atomic number and $e$ is the magnitude of the charge of one electron then the potential energy of the $\alpha$ particle at this distance from the nucleus is given by
$U=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\times \dfrac{(Ze)(2e)}{{{r}_{0}}}=\dfrac{2Z{{e}^{2}}}{4\pi {{\varepsilon }_{0}}{{r}_{0}}}$
If the initial kinetic energy of the $\alpha$ particle be$K$, then at the distance of closest approach
$\begin{aligned} & K=\dfrac{2Z{{e}^{2}}}{4\pi {{\varepsilon }_{0}}{{r}_{0}}} \\\ & {{r}_{0}}=\dfrac{2Z{{e}^{2}}}{4\pi {{\varepsilon }_{0}}K} \\\ \end{aligned}$
Putting the values of all the quantities we have
${{r}_{0}}=(9\times {{10}^{9}})\times \dfrac{2\times 50\times {{(1.6\times {{10}^{-19}})}^{2}}}{10\times {{10}^{6}}\times 1.6\times {{10}^{-19}}}=144\times {{10}^{-16}}=14.4\times {{10}^{-15}}m$
So the correct option is D.

Additional information:
If the initial kinetic energy of the $\alpha$ particle is very large, the particle will reach extremely close to the nucleus. In this condition, the nucleus will no longer be a point charge for the $\alpha$ particle, and Coulomb's law will no longer be applicable. $\alpha$ particle with energy greater than a certain limit will not be scattered back by the nucleus but the $\alpha$ particle will penetrate through it.

Note: To solve this kind of problem we need to remember the value of the quantity $\dfrac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}}$. We have to convert the energy given in $eV$ into $J$. Also, this type of problem can only be solved by applying Coulomb’s law if the energy of the $\alpha$ particle is less than some limiting value.