The smallest wavelength in the spectral lines of the Paschen series in the hydrogen atom is: (R is the Rydberg's constant in $A^{-1}$)
Answer & explanation
Correct answer: option 1
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: $\lambda_\text{min} = \frac{1}{R/9} = \frac{9}{R}$