The shortest wavelength in the Lyman series of hydrogen spectrum is 912 Å. The shortest wavelength present in Paschen series of spectral lines will be:

8208 Å

The correct answer is Option (1) → 8208 Å

Using Rydberg formula,

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

Lyman series, $n_1=1$

Paschen series, $n_2=3$

Shortest wavelength corresponds to $n_2=∞$

$\frac{1}{λ_{min}_L}=R_H\left(1-\frac{1}{∞}\right)=R_H=\frac{1}{912Å}$

$\frac{1}{λ_{min}_P}=R_H\left(\frac{1}{3^2}-\frac{1}{∞}\right)$

$⇒λ_{min}_P=9×912$

$=8208 Å$