Question

If $n_1$ and $n_2$ are the boundary value principle 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

$n_2=5$ & $n_1=2$

So longest wavelength means least energy difference transition i.e. $n_2=5to{n}_1=4$

$\frac{1}{\lambda }=R_H\left(1{)}^2\right(\frac{1}{4^2}-\frac{1}{5^2})$

so $\lambda =4\times {10}^{-6}m$