Question

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

• A. $6562.8\stackrel{\circ }{A}$
• B. $3281.4\stackrel{\circ }{A}$
• C. $6646.9\stackrel{\circ }{A}$
• D. $3580.2\stackrel{\circ }{A}$

Answer 1

The correct option is A $6562.8\stackrel{\circ }{A}$

We know,
$\frac{1}{\lambda }=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
where $\lambda =wavelengthofthetransition$
Given, $n_1=3and{n}_2=5,$
$\frac{1}{12818\stackrel{\circ }{A}}=R[\frac{1}{9}-\frac{1}{25}]=\frac{16R}{9\times 25}or,12818\stackrel{\circ }{A}=\frac{9\times 25}{16\times R}$ ...(i)
When $n_1=2and{n}_2=3$ $\frac{1}{\lambda }=R[\frac{1}{4}-\frac{1}{9}]=\frac{5R}{36}\lambda =\frac{36}{5R}$ ..(ii)
Dividing eq. (ii) by eq. (i)
$\frac{\lambda }{12818\stackrel{\circ }{A}}=\frac{36}{5R}\times \frac{16R}{9\times 25}=\frac{64}{125}\lambda =\frac{64}{125}\times 12818\stackrel{\circ }{A}=6562.8\stackrel{\circ }{A}$