Animated Solution for Physics - Atoms and Nuclei: A nucleus of mass $M$ emits $\gamma$-ray photon of frequency $\nu$. The loss of internal energy by the nucleus is [Take, $c$ is the speed of electromagnetic wave.]

When a nucleus undergoes radioactive decay and emits a gamma photon, it might seem like a simple transaction: the nucleus loses some internal energy, and that exact amount of energy flies away as the photon. But physics is rarely that simple. There is a hidden "tax" in this transaction, and it's paid in the form of recoil kinetic energy.

Let's break down the mechanics of this event. Initially, our nucleus of mass $M$ is sitting perfectly at rest. Its momentum is zero. Suddenly, it emits a gamma photon. Now, even though a photon has no rest mass, it carries momentum. According to the principles of quantum mechanics and relativity, the momentum of a photon is given by $p_\gamma = \frac{h u}{c}$, where $h u$ is its energy and $c$ is the speed of light.

Because the initial momentum of the system was zero, the final momentum must also be zero to satisfy the law of conservation of linear momentum. This means the nucleus cannot just sit there after firing off a photon. It must recoil in the exact opposite direction with a momentum equal in magnitude to the photon's momentum. Therefore, the recoil momentum of the nucleus is $p_{\text{recoil}} = \frac{h u}{c}$.

Now, a moving nucleus has kinetic energy. We can calculate this recoil kinetic energy using the classic relation between kinetic energy and momentum: $KE=\frac{p^2}{2M}$. Substituting our recoil momentum into this formula, we find that the kinetic energy of the recoiling nucleus is:

This brings us to the crux of the problem: the conservation of energy. The internal energy lost by the nucleus ($\Delta U$) doesn't just magically turn into the photon. It must supply the energy for the entire final state. This means the lost internal energy equals the energy of the gamma photon plus the kinetic energy of the recoiling nucleus.

So, we write our master equation:

Plugging in the values we've found, we get:

By factoring out $h u$, we arrive at our beautifully elegant final answer:

This result shows that the nucleus actually has to lose slightly more internal energy than just the energy of the photon it emits, simply because it has to "pay" for its own recoil!