Animated Solution for Chemistry - Atomic Structure: The ratio of the shortest wavelength of two spectral series of hydrogen spectrum is found to be about 9. The spectral series are

When scientists first observed the light emitted by hydrogen gas, they didn't see a continuous rainbow. Instead, they saw distinct, sharp lines of color. This was a profound mystery that eventually led to the birth of quantum mechanics. Niels Bohr proposed that electrons orbit the nucleus in specific, quantized energy levels. When an electron jumps from a higher energy level to a lower one, it sheds its excess energy by emitting a photon of light.

The energy of this photon dictates its wavelength, following the fundamental relation $E=\frac{hc}{\lambda }$. Because the energy levels are fixed, the emitted photons can only have specific wavelengths, creating what we call a spectral series. Each series is defined by its destination or "base" energy level ($n_1$). For example, all electrons falling down to $n=1$ create the Lyman series, while those falling to $n=2$ create the Balmer series.

In any given spectral series, there is a range of possible wavelengths. The longest wavelength occurs when the electron makes the smallest possible jump—from the very next energy level ($n_1+1$) down to the base level ($n_1$).

Conversely, the shortest wavelength corresponds to the most energetic photon. To release the maximum amount of energy, the electron must fall from the highest possible energy state. In an atom, the highest bound state is at an infinite distance from the nucleus, where the energy is zero. Therefore, the shortest wavelength in any series is produced by an electron transitioning from $n_2=\infty$ down to the base level $n_1$.

To calculate these wavelengths, we rely on the elegant Rydberg formula:

Here, $R_H$ is the Rydberg constant. Let's apply our condition for the shortest wavelength, where $n_2=\infty$. Since $\frac{1}{{\infty }^2}$ approaches zero, the formula simplifies beautifully:

Flipping this equation gives us a direct expression for the shortest wavelength of any series:

This tells us that the shortest wavelength is directly proportional to the square of the base energy level ($n_1^2$). Let's map this out for the classic hydrogen series: Lyman ($n_1=1$): ${\lambda }_L=\frac{1^2}{R_H}=\frac{1}{R_H}$ Balmer ($n_1=2$): ${\lambda }_B=\frac{2^2}{R_H}=\frac{4}{R_H}$ Paschen ($n_1=3$): ${\lambda }_P=\frac{3^2}{R_H}=\frac{9}{R_H}$ Brackett ($n_1=4$): ${\lambda }_{Bk}=\frac{4^2}{R_H}=\frac{16}{R_H}$ Pfund* ($n_1=5$): ${\lambda }_{Pf}=\frac{5^2}{R_H}=\frac{25}{R_H}$

The problem asks us to identify two spectral series whose shortest wavelengths have a ratio of exactly 9. Armed with our calculated values, this becomes a simple matching game. We need to find two wavelengths, ${\lambda }_A$ and ${\lambda }_B$, such that:

Looking at our list, if we take the shortest wavelength of the Paschen series ($\frac{9}{R_H}$) and divide it by the shortest wavelength of the Lyman series ($\frac{1}{R_H}$), the Rydberg constant $R_H$ perfectly cancels out:

This elegant integer ratio confirms that the two spectral series in question are indeed Paschen and Lyman. This problem beautifully illustrates how the quantized nature of the atom leads to simple, predictable mathematical relationships!