Animated Solution for Physics - Atoms and Nuclei: A free electron of $2.6\text{ eV}$ energy collides with a ${\text{H}}^{+}$ ion. This results in the formation of a hydrogen atom in the first excited state and a photon is released. Find the frequency of the emitted photon. ($h=6.6\times 10^{-34}\text{ Js}$)

Imagine a free electron zooming through space with a kinetic energy of $2.6\text{ eV}$. It encounters a stationary ${\text{H}}^{+}$ ion—a bare proton waiting to capture an electron.

When the electron is captured by the proton, it transitions from being a "free" particle to a "bound" particle. This fascinating process is known as radiative recombination. The universe demands that energy be conserved, so the excess energy from this capture must go somewhere. It is released into the universe as a brilliant flash of light—a photon!

To find the energy of the emitted photon, we must first understand the initial and final energy states of our electron.

Initially, the electron is free. It is not bound by any electrostatic forces, so its potential energy is effectively zero. Its total energy is simply its kinetic energy:

After the collision, the electron is captured into the first excited state of the newly formed hydrogen atom. A common trap here is assuming the first excited state means $n=1$. Remember, $n=1$ is the ground state! The first excited state corresponds to $n=2$.

Using Bohr's model, the energy of an electron in the $n^{\text{th}}$ orbit of a hydrogen atom is given by:

Substituting $n=2$, we find the final energy of the electron:

Energy Conservation

The electron has dropped from an energy of $+2.6\text{ eV}$ down to $-3.4\text{ eV}$. The energy difference is carried away by the emitted photon. By the principle of conservation of energy:

Let's substitute our values carefully, watching out for the negative signs:

So, the emitted photon carries exactly $6.0\text{ eV}$ of energy.

Finding the Frequency

We know the energy of the photon, but the question asks for its frequency. The relationship between a photon's energy and its frequency is given by the famous Planck-Einstein relation:

Here is where many students make a silly mistake. Our energy is in electron volts ($\text{eV}$), but Planck's constant ($h$) is given in standard SI units ($\text{J}\cdot \text{s}$). We must convert the energy into Joules before proceeding!

Now, we can solve for the frequency $ u$:

The options provided are in Megahertz ($\text{MHz}$). Since $1\text{ MHz}=10^6\text{ Hz}$, we divide our result by $10^6$:

This perfectly matches option (c). The elegance of energy conservation allows us to seamlessly connect the macroscopic kinetic energy of a free electron to the quantum frequency of an emitted photon!