Animated Solution for Physics - Atoms and Nuclei: In the line spectra of hydrogen atom, difference between the largest and the shortest wavelengths of the Lyman series is $304\text{ A˚}$. The corresponding difference for the Paschan series (in $\text{A˚}$) is ......... .

Welcome to an exciting journey into the quantum world of the hydrogen atom! Today, we are going to decode the spectral lines emitted when an electron dances between different energy levels. This problem is a classic application of the Bohr model and the Rydberg formula, testing your ability to connect abstract energy transitions to measurable wavelengths.

Imagine an electron in a hydrogen atom. When it jumps from a higher energy level to a lower one, it releases energy in the form of a photon. The wavelength of this photon depends on the specific energy levels involved.

The problem gives us information about the Lyman series and asks us to find a corresponding value for the Paschen series. The Lyman series occurs when an electron falls to the ground state ($n_f=1$). The Paschen series occurs when it falls to the second excited state ($n_f=3$).

To find the wavelengths, we rely on the powerful Rydberg formula. This equation beautifully links the wavelength of the emitted photon to the initial and final energy levels of the electron.

Here, $R$ is the Rydberg constant, $n_f$ is the final energy level, and $n_i$ is the initial energy level. We will use this master equation to analyze both the Lyman and Paschen series.

Let's start with the Lyman series, where $n_f=1$. The longest wavelength corresponds to the smallest energy jump, which is from $n_i=2$ to $n_f=1$.

Conversely, the shortest wavelength corresponds to the largest energy jump, which is from $n_i=\infty$ to $n_f=1$.

We are given that the difference between these two wavelengths is $304\text{ A˚}$. Let's set up the equation to find the value of $1/R$.

Since $\Delta \lambda =304$, we have $\frac{1}{3R}=304$, which means $\frac{1}{R}=3\times 304=912$. We will keep this value handy for the next phase.

Now, let's shift our focus to the Paschen series, where $n_f=3$. The longest wavelength here is for the transition from $n_i=4$ to $n_f=3$.

The shortest wavelength for the Paschen series is for the transition from $n_i=\infty$ to $n_f=3$.

We need to find the difference between these two wavelengths, let's call it $\Delta {\lambda }^{\prime }$.

We are almost there! We have the expression for the difference in the Paschen series, and we already found the value of $1/R$ from the Lyman series. Let's substitute it in.

Now, it's just a matter of simple arithmetic. Multiplying $81$ by $912$ and dividing by $7$ gives us approximately $10553.14$.

And there we have it! The corresponding difference for the Paschen series is $10553$. This problem beautifully demonstrates how the spectral series are interconnected through the Rydberg constant.