The smallest wavelength in the spectral lines of the Paschen series in the hydrogen atom is: (R is the Rydberg's constant in $A^{-1}$)

$\frac{9}{R}Å$

The correct answer is Option (1) → $\frac{9}{R}Å$

The Paschen series corresponds to transitions to $n = 3$ level in hydrogen.

Smallest wavelength corresponds to the largest energy difference → transition from $n = \infty$ to $n = 3$.

Wavelength formula (Rydberg formula):

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

For $n \to \infty$:

$\frac{1}{\lambda_\text{min}} = R \left( \frac{1}{3^2} - 0 \right) = \frac{R}{9}$

Final Answer: $\lambda_\text{min} = \frac{1}{R/9} = \frac{9}{R}$