Question

If $n_1$ and $n_2$ are the boundary value principal quantum numbers of a portion of spectrum of emission spectrum of H atom, determine the wavelength (in metre) corresponding to last line (longest $\lambda$). Given : $n_1+n_2=7,n_2-n_1=3,and{R}_H=1.097\times {10}^7{m}^{-1}.$ (Give your answer in multiple of ${10}^{-6}$)

Answer 1

Here $n_2=5and{n}_1=2$
So longest wavelength means least energy difference transition i.e. $n_2=5to{n}_1=4$
$\frac{1}{\lambda }=R_H(1{)}^2(\frac{1}{(4{)}^2}-\frac{1}{(5{)}^2})$
So $\lambda =4\times {10}^{-6}$ m