Question

Consider the following two electronic transition possibilities in a hydrogen atom as pictured below:
(a) The electron drops from third Bohr orbit to second Bohr orbit followed with the next transition from second to first Bohr orbit.
(b) The electron drops from third Bohr orbit to first Bohr orbit directly. Show that the sum of energies for the transitions $n=3$ to $n=2$ and $n=2$ to $n=1$ is equal to the energy of transition for $n=3$ to $n=1$.

Answer 1

Applying, $\Delta E=R_H[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
For $n=3$ to $n=2$;
$\Delta E_{3\to 2}=R_H[\frac{1}{2^2}-\frac{1}{3^2}]=R_H\times \frac{5}{36}$ ...(i)
For $n=2$ to $n=1$;
$\Delta E_{2\to 1}=R_H[\frac{1}{1^2}-\frac{1}{2^2}]=R_H\times \frac{3}{4}$ ....(ii)
For $n=3$ to $n=1$;
$\Delta E_{3\to 1}=R_H[\frac{1}{1^2}-\frac{1}{3^2}]=R_H\times \frac{8}{9}$ ...(iii)
Adding equations (i) and (ii),
$R_H(\frac{5}{36}+\frac{3}{4})=R_H\left(\frac{5+27}{36}\right)=R_H\times \frac{8}{9}$
Thus, $\Delta E_{3\to 1}=\Delta E_{3\to 2}+\Delta E_{2\to 1}$.