If the minimum wavelength corresponding the Paschen series of hydroge…

The correct answer is Option (2) → 364.4 nm

Wavelength of radiation emitted in hydrogen spectra

$\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

For Paschen $n_1=3, n_2 \geq 4$

for $\lambda_{\min} n_1=3, n_2=\infty$

$\frac{1}{\lambda_1}=\frac{R}{9}$

For Balmer $n_1=2, n_2 \geq 3$

For $\lambda_{\min } n_1=2, n=\infty$

$\frac{1}{\lambda_2}=\frac{R}{4}$

$\frac{\lambda_2}{\lambda_1}=\frac{4}{9}$

$\lambda_2=\frac{4}{9} \times \lambda_1$

$=\frac{4}{9} \times 820$

$\approx 364.4 ~nm$