Animated Solution for Physics - Atoms and Nuclei: The electron in a hydrogen atom first jumps from the third excited state to the second excited state and subsequently to the first excited state. The ratio of the respective wavelengths ${\lambda }_1/{\lambda }_2$ of the photons emitted in this process is

Imagine a staircase, but not an ordinary one. In this quantum staircase, the steps are not evenly spaced. As you climb higher, the steps get closer and closer together. This is exactly what the energy levels of a hydrogen atom look like. The ground floor is our $n=1$ state. When an electron gains energy, it climbs up. The first step up is the first excited state ($n=2$), the second step is the second excited state ($n=3$), and so on.

In our problem, the electron starts its journey on the third excited state. Let's decode this terminology. If the ground state is $n=1$, then the first excited state is $n=2$, the second is $n=3$, and the third excited state is $n=4$. So, our electron is perched up on the $n=4$ step, ready to make a move.

The electron doesn't jump all the way down at once. It decides to take a pit stop. It jumps from the third excited state ($n=4$) to the second excited state ($n=3$).

When an electron drops to a lower energy level, it has to shed its excess energy. It does this by throwing out a tiny packet of light called a photon. The wavelength of this photon, let's call it ${\lambda }_1$, is dictated by the energy difference between the two steps.

To find this wavelength, we bring out our trusty tool: the Rydberg formula.

Here, $R$ is the Rydberg constant, $n_i$ is the initial state, and $n_f$ is the final state. For our first leap, $n_i=4$ and $n_f=3$. Let's plug these in:

To subtract these fractions, we need a common denominator, which is $9\times 16=144$.

We'll keep it in this reciprocal form for now. It will make our lives much easier later!

Now, the electron is at $n=3$. But it's not done yet. It takes another jump, this time down to the first excited state ($n=2$).

Again, it sheds energy by emitting a second photon, with a wavelength we'll call ${\lambda }_2$. Let's use the Rydberg formula again. This time, $n_i=3$ and $n_f=2$.

The common denominator here is $4\times 9=36$.

An Inverse Dance

It's crucial to remember the relationship between the energy of a photon and its wavelength. According to the Planck-Einstein relation, $E=\frac{hc}{\lambda }$. This means energy and wavelength are inversely proportional. A larger energy jump produces a photon with a shorter wavelength, and a smaller energy jump produces a photon with a longer wavelength.

When the electron jumps from $n=4$ to $n=3$, the energy levels are relatively close together. When it jumps from $n=3$ to $n=2$, the energy gap is significantly larger. Therefore, we should expect the energy of the second photon to be greater than the first, which means ${\lambda }_2$ should be smaller than ${\lambda }_1$. Consequently, the ratio $\frac{{\lambda }_1}{{\lambda }_2}$ should be greater than 1. Keeping this physical intuition in mind is a great way to sanity-check our final mathematical result!

Finding the Ratio

The question asks for the ratio of the two wavelengths, $\frac{{\lambda }_1}{{\lambda }_2}$. We have expressions for $\frac{1}{{\lambda }_1}$ and $\frac{1}{{\lambda }_2}$. A clever algebraic trick is to realize that:

This is brilliant because we can just plug in our fractions directly!

When we divide by a fraction, we multiply by its reciprocal. And look at the Rydberg constant $R$—it's in both the numerator and the denominator, so it gracefully cancels out!

Now, it's just a matter of simple arithmetic. We know that $36\times 4=144$, so we can simplify the fraction:

And there we have it! The ratio of the wavelengths is $20/7$. This problem beautifully illustrates how the discrete, quantized nature of energy levels directly translates into specific, predictable wavelengths of light. Every jump tells a mathematical story!