Question

calculate the wavelength of the second line of lyman series in hydrogen spectra

Answer 1

Answer of this question is

Answer 2

Answer: $1.026\times 10^{-7}m$

Explanation:

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

Where,

$\lambda$ = Wavelength of radiation = ?

$R_H$ = Rydberg's Constant =$1.09737\times 10^7m^{-1}$

$n_f$ = Higher energy level = 3 (second line of Lyman series)

$n_i$ = Lower energy level = 1

Putting the values, in above equation, we get

$\frac{1}{\lambda}=1.09737\times 10^7m^{-1}\left(\frac{1}{1^2}-\frac{1}{3^2} \right)$

$\frac{1}{\lambda}=0.975\times 10^7m$

$\lambda=1.026\times 10^{-7}m$