Animated Solution for Chemistry - Atomic Structure: For any given series of spectral lines of atomic hydrogen, let $\Delta \bar{\nu }={\bar{\nu }}_{\max }-{\bar{\nu }}_{\min }$ be the difference in maximum and minimum frequencies in ${\text{cm}}^{-1}$. The ratio $\Delta {\bar{\nu }}_{\text{Lyman}}/\Delta {\bar{\nu }}_{\text{Balmer}}$ is

Imagine you are looking at the energy levels of a hydrogen atom. The problem asks us to find the ratio of the difference between the maximum and minimum wave numbers (or frequencies in ${\text{cm}}^{-1}$) for the Lyman and Balmer series.

First, let's establish what wave number ($\bar{ u}$) actually represents. The wave number is simply the reciprocal of the wavelength ($\lambda$), and it is directly proportional to the energy difference ($\Delta E$) of the transition.

Therefore, the maximum wave number ($\bar{ u}_{\max}$) corresponds to the maximum energy transition. For any series ending at a base level $n_1$, this happens when the electron falls from the highest possible state, which is $n_2=\infty$. Conversely, the minimum wave number ($\bar{ u}_{\min}$) corresponds to the smallest energy jump, which occurs when the electron falls from the immediately adjacent higher level, $n_2=n_1+1$.

Rydberg Formula

To calculate these wave numbers, we rely on the famous Rydberg formula:

Here, $R_H$ is the Rydberg constant, $n_1$ is the lower energy level (the destination), and $n_2$ is the higher energy level (the origin).

Let's focus on the Lyman series first. For the Lyman series, the base level is always $n_1=1$.

For the maximum wave number, the electron falls from $n_2=\infty$:

For the minimum wave number, the electron falls from the very next level, $n_2=2$:

Now, we find the difference for the Lyman series:

Now, let's shift our attention to the Balmer series. Here, the base level is $n_1=2$.

For the maximum wave number, the transition is from $n_2=\infty$ to $n_1=2$:

For the minimum wave number, the transition is from the next level, $n_2=3$ to $n_1=2$:

Let's find the difference for the Balmer series. Be careful with the LCM here!

Finally, we take the ratio of the two differences we just calculated:

The $R_H$ terms cancel out beautifully, leaving us with:

So, the required ratio is 9:4, which matches option (c).