Question

Light of wavelength $12818\mathring{A}$ is emitted when the electron of a hydrogen atom drops from $5$th to $3$rd orbit. Find the wavelength of the photon emitted when the electron falls from $3$rd to $2$nd orbit.

Answer 1

We know that,
$\frac{1}{\lambda }=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
When, $n_1=3$ and $n_2=5$,
$\frac{1}{12818}=R[\frac{1}{9}-\frac{1}{25}]=\frac{16R}{9\times 25}$
or $12818=\frac{9\times 25}{16\times R}$ ...(i)
When, $n_1=2$ and $n_2=3$,
$\frac{1}{\lambda }=R[\frac{1}{4}-\frac{1}{9}]=\frac{5R}{36}$
$\lambda =\frac{36}{5R}$ ....(ii)
Dividing equation (ii) by equation (i),
$\frac{\lambda }{12818}=\frac{36}{5R}\times \frac{16R}{9\times 25}=\frac{64}{125}$
$\lambda =\frac{64}{125}\times 12818=6562.8\mathring{A}$.