Animated Solution for Physics - Atoms and Nuclei: The recoil speed of a hydrogen atom after it emits a photon in going from $n=5$ state to $n=1$ state is ........ m/s.

Imagine a hydrogen atom floating perfectly still in the vast emptiness of space. Suddenly, an electron that was orbiting in the highly excited $n=5$ state drops down to the ground state, $n=1$. This massive drop in energy cannot just disappear; it is released as a brilliant burst of electromagnetic radiation—a photon.

But physics demands balance. If a photon shoots off in one direction carrying momentum, the atom cannot simply remain stationary. Just as a cannon recoils backward when it fires a heavy cannonball, the hydrogen atom must recoil in the opposite direction to conserve the total momentum of the system. This beautiful interplay between quantum energy transitions and classical mechanics is the heart of our problem.

Before the emission, the atom is at rest, meaning the total initial momentum of the system is zero. According to the Law of Conservation of Linear Momentum, the total final momentum must also be zero.

Mathematically, this means:

This implies that the magnitude of the atom's recoil momentum must exactly equal the momentum of the emitted photon:

We know that the momentum of a classical particle is $p=mv$. But what about a photon, which has no mass? Thanks to Einstein, we know that a photon's momentum is directly related to its energy by the simple equation $p=\frac{\Delta E}{c}$, where $c$ is the speed of light. Equating the two gives us our master equation:

To find the recoil velocity $v$, we first need to determine the energy of the emitted photon, $\Delta E$. We turn to the Bohr model of the hydrogen atom. The energy of an electron in the $n^{\text{th}}$ orbit is given by $E_n=-\frac{13.6}{n^2}\text{ eV}$.

The energy difference between the $n=5$ and $n=1$ states is:

Let's compute this step-by-step:

While electron-volts are incredibly convenient for atomic physics, our momentum equation requires standard SI units (Joules, kilograms, meters per second) to yield a velocity in $\text{m/s}$. We must convert the energy into Joules by multiplying by the elementary charge ($1.6\times 10^{-19}\text{ C}$):

Now, we return to our master equation, $v=\frac{\Delta E}{mc}$. We need the mass of the hydrogen atom. Since a hydrogen atom is essentially just a proton and an electron, and the electron's mass is negligible, we use the mass of a proton: $m=1.67\times 10^{-27}\text{ kg}$.

Plugging in all our values:

The hydrogen atom recoils with a speed of $4.17\text{ m/s}$. It is a small, gentle drift compared to the speed of light, but it is a very real physical consequence of emitting a photon!

Does Recoil Change the Photon's Color?

Here is a fascinating thought experiment: Since the recoiling atom gains some kinetic energy ($\frac{1}{2}m{v}^2$), where does that energy come from? It must come from the total energy released in the transition. This means the emitted photon actually has slightly less energy than the exact difference between the atomic energy levels!

This tiny loss of energy causes a minuscule redshift in the photon's wavelength. For a light atom like hydrogen emitting visible or ultraviolet light, this shift is practically negligible. However, when heavy nuclei emit highly energetic gamma rays, this recoil effect becomes incredibly important. Overcoming this recoil energy loss is the foundational principle behind the famous Mössbauer effect, a discovery that earned a Nobel Prize. Physics is beautifully interconnected!