Correct Option: (A) infrared region

Step 1 (Governing Law & Formula): [0.5 Mark]

The spectral series of a hydrogen atom is governed by the Rydberg formula, which determines the wavelength ($\lambda$) of the emitted radiation when an electron transitions from a higher energy level ($n_i$) to a lower energy level ($n_f$):

$$\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$$

Where $R$ is the Rydberg constant. For the Paschen series, the electron transitions to the third energy level, hence $n_f = 3$ and $n_i = 4, 5, 6, \dots$

Step 2 (Step-by-Step Mathematical Working): [0.25 Mark]

Substituting the value of $n_f$ for the Paschen series into the formula:

$$\frac{1}{\lambda} = R \left( \frac{1}{3^2} - \frac{1}{n_i^2} \right) = R \left( \frac{1}{9} - \frac{1}{n_i^2} \right)$$

Since $n_i > 3$, the energy difference $\Delta E = \frac{hc}{\lambda}$ is relatively small compared to transitions ending at $n_f = 1$ (Lyman) or $n_f = 2$ (Balmer). Lower energy transitions correspond to longer wavelengths, which fall within the infrared portion of the electromagnetic spectrum.

Step 3 (Final Concluded Answer): [0.25 Mark]

Based on the calculated energy range and wavelength values, the Paschen series lies entirely in the infrared region.

💡 EL10 Concept Insight:

Think of the electron like a ball bouncing down stairs; the bigger the jump, the more energy it releases. Because the Paschen series involves "smaller" jumps down to the 3rd step, it releases less energy than the big leaps to the 1st or 2nd steps, and that lower energy shows up as infrared light instead of visible or ultraviolet light.