Question

A nucleus of mass number $220$ decays by decay. The energy released in the reaction is $5MeV$. The kinetic energy of an $\alpha$ particle is:
(A) $\dfrac{1}{54}MeV$
(B) $\dfrac{27}{11}MeV$
(C) $\dfrac{54}{11}MeV$
(D) $\dfrac{55}{54}MeV$

Answer 1

In the given question, we have been given some information about a radioactive element. We have been provided with its mass number and we have been told that it undergoes alpha decay; we have also been given the total energy released in the reaction and we have been asked the kinetic energy of the alpha particle. At first glance, we know nothing about the alpha particle. But we are aware of its composition, that it has two electrons and two protons. To find the distribution of the total energy in a radioactive reaction, all we need is the mass number of the reactants and the products. Let’s see the detailed solution.

Formula Used ${{K}_{\alpha }}=\dfrac{{{M}_{d}}}{{{M}_{\alpha }}+{{M}_{d}}}\times E$

Complete step by step answer:
A radioactive substance is said to have undergone alpha decay when it emits an alpha particle and turns into a daughter nucleus. The alpha nucleus has two protons and two electrons, as mentioned above. So the loss of an alpha particle from the mother nucleus will reduce its atomic number and mass number by two and four respectively.
The process of alpha decay can be understood with the help of the following equation:
${}_{Z}^{220}A\to {}_{2}^{4}He+{}_{Z-2}^{216}B$ where ${}_{Z}^{220}A$ denotes the mother nucleus, ${}_{2}^{4}He$ represents the alpha particle and ${}_{Z-2}^{216}B$ denotes the daughter nucleus (the superscripts and the subscripts denote the mass number and the atomic number of the elements respectively).
We know that for a pair of daughter nucleus and alpha particle, the kinetic energy of the alpha particle can be given as
${{K}_{\alpha }}=\dfrac{{{M}_{d}}}{{{M}_{\alpha }}+{{M}_{d}}}\times E$ where ${{M}_{d}}$ is the mass number of the daughter nucleus, ${{M}_{\alpha }}$ is the mass of the alpha particle and $E$ is the total energy of the reaction.
Substituting the values in the above equation, we get

& {{K}_{\alpha }}=\dfrac{216}{4+216}\times 5MeV \\\ & \Rightarrow {{K}_{\alpha }}=\dfrac{216}{220}\times 5MeV \\\ & \Rightarrow {{K}_{\alpha }}=\dfrac{54}{11}MeV \\\ \end{aligned}$$ **Hence we can say that option (C) is the correct answer.** **Note** The expression for the division of the total energy comes from the principle of momentum conservation applied on the alpha particle and the daughter nucleus. But since we have been asked to find the kinetic energy of the alpha particle, deriving the expression in the solution would take away our solving time. You can apply the expression or the formula directly as long as you are aware of the concept behind it.