Animated Solution for Physics - Atoms and Nuclei: A nucleus with mass number 220 initially at rest emits an $\alpha$-particle. If the $Q$ value of the reaction is 5.5 MeV, calculate the kinetic energy of the $\alpha$-particle

Imagine a heavy, unstable nucleus, sitting perfectly still. It's like a coiled spring, packed with immense nuclear energy, just waiting for the right moment to release it. In this problem, we have a nucleus with a mass number of $A=220$. Suddenly, it undergoes alpha decay, ejecting an $\alpha$-particle and transforming into a slightly lighter daughter nucleus.

This explosive event releases a total energy known as the $Q$-value, which is given as $5.5\text{ MeV}$. But how is this energy shared between the two newly formed particles? Does it split equally? To answer this, we need to invoke two of the most powerful laws in physics: the conservation of energy and the conservation of momentum.

First, let's look at the conservation of energy. The total energy released must equal the sum of the kinetic energies of the products. If we let $K_{\alpha }$ be the kinetic energy of the $\alpha$-particle and $K_D$ be the kinetic energy of the daughter nucleus, we can write:

Now, let's bring in the conservation of momentum. Before the decay, the parent nucleus was at rest, meaning its initial momentum was zero. For the total momentum to remain zero after the decay, the $\alpha$-particle and the daughter nucleus must fly apart in exactly opposite directions with equal magnitudes of momentum.

Let $p$ be the magnitude of momentum for both particles. We know the relationship between kinetic energy, momentum, and mass is given by:

Since $p$ is the same for both particles, their kinetic energy is inversely proportional to their mass. This is a crucial insight! The lighter particle will zip away much faster and carry the lion's share of the kinetic energy.

Let's set up the ratio of their kinetic energies. The mass of the $\alpha$-particle is $4\text{ amu}$, and the mass of the daughter nucleus is $220-4=216\text{ amu}$.

This tells us that the heavy daughter nucleus only gets $\frac{1}{54}$ of the kinetic energy that the $\alpha$-particle gets. We can express $K_D$ in terms of $K_{\alpha }$:

Now, we substitute this back into our energy conservation equation:

Factoring out $K_{\alpha }$, we get:

Solving for $K_{\alpha }$, we multiply both sides by $\frac{54}{55}$:

And there we have it! The $\alpha$-particle walks away with a whopping $5.4\text{ MeV}$ of the total $5.5\text{ MeV}$ released.

For competitive exams like JEE, time is of the essence. You can bypass the derivation by remembering a beautiful general formula for the kinetic energy of an $\alpha$-particle emitted from a nucleus at rest:

Let's test it with our numbers:

It works perfectly! Understanding the physics behind the formula not only builds your intuition but also gives you a reliable tool to crush these problems in seconds. Keep visualizing the physics, and the math will naturally follow!