Question

Consider the following two electronic transition possibilities in hydrogen atom as:
$\left(1\right)$ The electron drops from third Bohr's orbit to second Bohr's orbit followed with the next transition from second to first Bohr's orbit.
$\left(2\right)$ The electron drops from third Bohr's orbit to first Bohr's orbit directly.
Then the statement :
"The sum of the energies for the transitions $n=3$ to $n=2$ and $n=2$ to $n=1$ = the energy of transition for $n=3$ to $n=1$."
Enter $1$ if true, else enter $0$.

Answer 1

$\Delta E=R_H[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
For $3$ to $2$ $\Delta E_{3\to 2}=R_H[\frac{1}{2^2}-\frac{1}{3^2}]$
For $2$ to $1$ $\Delta E_{2\to 1}=R_H[\frac{1}{1^2}-\frac{1}{2^2}]$
For $3$ to $1$ $\Delta E_{3\to 1}=R_H[\frac{1}{1^2}-\frac{1}{3^2}]$
It is evident from Eqs. (i),(ii) and (iii), that $\Delta E_{3\to 1}=\Delta E_{3\to 2}+\Delta E_{2\to 1}$
(b) Also,$E=hv$; thus frequencies are also additive.
But $E=\frac{hc}{\lambda }$ and thus wavelength are not additive.