Question

Calculate the shortest and longest wavelength of Balmer series of $H_2$ spectrum. Given $R=1.097\times {10}^7$ $m^{-1}$.

Answer 1

$Forshortestwavelength,n_1=1\&n_2=\infty Now,\frac{1}{\lambda }=R[\frac{1}{{\left(n_1\right)}^2}-\frac{1}{{{(n}_2)}^2}]\frac{1}{\lambda }=1.097\times {10}^7[\frac{1}{1^2}-\frac{1}{{\infty }^2}]\frac{1}{\lambda }=1.097\times {10}^7\lambda =911.6\mathring{A}Forlongestwavelength,n_1=1\&n_2=2So,\frac{1}{\lambda }=1.097\times {10}^7[\frac{1}{1^2}-\frac{1}{2^2}]\frac{1}{\lambda }=1.097\times {10}^7\times \frac{3}{4}d\frac{1}{\lambda }=0.823\times {10}^7\lambda =1215.06\mathring{A}$