Animated Solution for Physics - Atoms and Nuclei: Taking the wavelength of first Balmer line in hydrogen spectrum ($n=3$ to $n=2$) as $660\text{ nm}$, the wavelength of the $2^{\text{nd}}$ Balmer line ($n=4$ to $n=2$) will be

The study of atomic spectra is one of the most fascinating chapters in modern physics. It bridges the gap between the macroscopic world we see and the quantum world that governs it. In this problem, we are exploring the Balmer series of the hydrogen atom, a classic example of quantum jumps.

Imagine you are looking at the energy levels of a hydrogen atom. The Balmer series is special because it corresponds to all the electron transitions that end exactly at the second energy level ($n=2$).

The first line of this series, often called the H-alpha line, happens when an electron drops from the very next level, $n=3$, down to $n=2$. The problem tells us that this specific jump releases a photon with a wavelength of $660\text{ nm}$.

Our goal is to find the wavelength of the second line in this series, which occurs when an electron takes a slightly larger leap from $n=4$ down to $n=2$.

To solve this, we need our trusty tool: the Rydberg formula. This elegant equation relates the wavelength of the emitted light to the energy levels involved in the transition.

Here, $R$ is the Rydberg constant, $Z$ is the atomic number (which is 1 for hydrogen), $n_f$ is the final energy level, and $n_i$ is the initial energy level. Because $R$ and $Z$ are constants for our specific atom, we can simplify our lives by writing this as a proportionality:

Let's apply this to our two scenarios. For the first Balmer line, the electron jumps from $n_i=3$ to $n_f=2$. Plugging these into our proportionality gives:

This is our first crucial piece of the puzzle. Now, let's look at the second Balmer line, where the jump is from $n_i=4$ to $n_f=2$. We'll call its unknown wavelength $\lambda$.

Now, we could try to find the exact value of the proportionality constant, but that's a trap! It takes too much time. Instead, we use a classic physicist's trick: taking the ratio. By dividing our first equation by our second equation, the constants completely vanish.

Let's simplify that complex fraction by multiplying by the reciprocal:

All that's left is a simple multiplication to isolate our unknown wavelength.

And there we have it! The wavelength of the second Balmer line is $488.9\text{ nm}$. Notice how the larger energy jump from $n=4$ resulted in a shorter wavelength compared to the jump from $n=3$. This perfectly aligns with the principle that higher energy photons have shorter wavelengths.