The shortest wavelength present in the Lyman series of spectral lines is:

(Given Rydbage constant $R=1.097 \times 10^7 m^{-1}$ )

$9.1 \times 10^{-8} m$

The correct answer is Option (1) → $9.1 \times 10^{-8} m$

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

$n_f=1$ [for Lyman series]

$n_i=∞$ [for shortest wavelength]

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

$⇒λ=R_H≃9.1 \times 10^{-8} m$