Animated Solution for Physics - Atoms and Nuclei: The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 \AA. The wavelength of the second spectral line in the Balmer series of singly ionized helium atom is

Imagine looking at a neon sign or the glowing gas in a discharge tube. The beautiful colors you see aren't just random light; they are the specific "fingerprints" of the atoms inside. According to Bohr's atomic model, electrons live in strictly quantized energy levels. When an electron gets excited and then falls back down to a lower energy level, it releases a photon of light. The energy of this photon—and therefore its color or wavelength—corresponds exactly to the energy difference between those two levels.

In the vast spectrum of hydrogen, the Balmer series is particularly special because it's the only series that falls in the visible region of light. By definition, any electron that drops down to the second energy level ($n_f=2$) produces a Balmer line.

But what do we mean by the "first" and "second" lines?

- The first line corresponds to the smallest possible jump, which is from $n_i=3$ down to $n_f=2$. A smaller energy jump means a lower frequency and a longer wavelength. - The second line corresponds to the next possible jump, from $n_i=4$ down to $n_f=2$. This is a larger energy jump, resulting in a shorter wavelength.

Rydberg's Formula

To calculate the exact wavelengths of these spectral lines, we rely on the elegant Rydberg's Formula:

Here, $\lambda$ is the wavelength, $R$ is the Rydberg constant, $Z$ is the atomic number of the nucleus, and $n_i$ and $n_f$ are the initial and final energy levels. Notice the crucial $Z^2$ term! This tells us that as the nucleus gets heavier (more protons), the energy levels are pulled much tighter together, drastically changing the emitted wavelengths.

Let's apply this to the first part of our problem: the first line of the hydrogen atom. For hydrogen, the atomic number is $Z_1=1$. The first line of the Balmer series means the transition is from $n_i=3$ to $n_f=2$.

Plugging this into our formula:

Now, let's look at the singly ionized helium atom ($H{e}^{+}$). Because it has lost one electron, it behaves just like hydrogen, allowing us to use the same formula. However, helium has two protons, so $Z_2=2$.

The problem asks for the second line of the Balmer series, which means the transition is from $n_i=4$ to $n_f=2$.

Setting up the equation:

We have two equations and we know ${\lambda }_1=6561\text{ A˚}$. We need to find ${\lambda }_2$. Instead of plugging in the messy value of the Rydberg constant $R$, we can use a powerful physicist's trick: taking the ratio.

By dividing the equation for ${\lambda }_1$ by the equation for ${\lambda }_2$, the constant $R$ beautifully cancels out:

Now, it's just a matter of careful fraction arithmetic. Let's substitute the known values:

Let's solve the numerators and denominators:

Flipping and multiplying the bottom fraction:

Finally, we isolate ${\lambda }_2$:

This method of using ratios is incredibly versatile. Whether you are comparing different spectral series (like Lyman vs. Paschen) or different hydrogen-like ions (like $L{i}^{2+}$ or $B{e}^{3+}$), setting up the two states and dividing them will almost always lead you to the cleanest, fastest solution. Keep this tool sharp in your mathematical toolkit!