Animated Solution for Physics - Atoms and Nuclei: A nucleus with mass number $184$ initially at rest emits an $\alpha$-particle. If the $Q$-value of the reaction is $5.5\text{ MeV}$, calculate the kinetic energy of the $\alpha$-particle.

It has a mass number of $184$. It's unstable and about to undergo alpha decay.

Boom! It emits an $\alpha$-particle. The original nucleus transforms into a new nucleus, $Y$, and recoils backwards. Because the initial momentum was zero, the final momenta of $Y$ and the $\alpha$-particle must be equal in magnitude and opposite in direction to conserve momentum. This is a fundamental principle of physics that governs all such decays.

Since their momenta are equal in magnitude ($p_Y=p_{\alpha }$), we can equate their kinetic energies in terms of their masses.

This gives us the powerful relation:

Let's plug in the masses. The mass of the $\alpha$-particle is approximately $4\text{ amu}$, and the remaining nucleus $Y$ has a mass of $184-4=180\text{ amu}$. Rearranging this, we find that the kinetic energy of nucleus $Y$ is just a fraction of the kinetic energy of the $\alpha$-particle:

This $Q$-value is exactly the total kinetic energy shared between the $\alpha$-particle and the recoiling nucleus $Y$.

We can substitute the kinetic energy of $Y$ with the kinetic energy of the $\alpha$-particle divided by $45$. This gives us an equation entirely in terms of the $\alpha$-particle's kinetic energy:

Taking the common denominator, we get:

Solving for the kinetic energy of the $\alpha$-particle gives us:

And there we have it! The $\alpha$-particle carries away the lion's share of the energy ($5.38\text{ MeV}$) because it is much lighter than the recoiling nucleus. The correct option is (d).