Question

The wavelength(in nm) of the radiation corresponding to the spectral line of the lowest frequency in Lyman series in the spectrum of hydrogen atom will be:
.(R${}_H$ = 1.9678 $\times$ 10${}^7$ m${}^{-1}$ c = 3 $\times$ 10${}^8$ m s${}^{-1}$, h = 6.625 $\times$ 10${}^{-34}$ J s)

• A. 121.57
• B. 102.57
• C. 1215.7
• D. 1025.7

Answer 1

Answer: (A)

121.57

$\frac{1}{\lambda }=R(\frac{1}{n_1^2}-\frac{1}{n_2^2})=1.09678\times {10}^7\times (\frac{1}{1^2}-\frac{1}{2^2})=82258.5/cm$
$\lambda =\frac{1}{82258.5}=121.57\times {10}^{-7}$ cm or 121.57 nm