The shortest wavelength present in the Lyman series of spectral lines is:
(Given Rydbage constant $R=1.097 \times 10^7 m^{-1}$ )
Answer & explanation
Correct answer: option 1
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]
$∴\frac{1}{λ}=R_H\left(\frac{1}{1^2}-\frac{1}{∞}\right)$
$⇒λ=R_H≃9.1 \times 10^{-8} m$