Question

Hydrogen atoms are excited from ground state. Its spectrum contains wavelength $486nm$. Find, what transition does the line corresponds to. Also find from this information what other wavelengths will be present in the spectrum?

Answer 1

Wavelength $486nm$, i.e., $4860\mathring{A}$ indicates that the spectrum is in visible region, i.e., Balmer series.

$\frac{1}{\lambda }=R{Z}^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

$\frac{1}{4860\times {10}^{-8}}=109677.76\times 1^2[\frac{1}{2^2}-\frac{1}{n_2^2}]$

On solving, we get;

$n_2^2=16$

$n_2=4$

Thus, transition is from, $4\to 2$

Other transitions in the spectrum are

$4\to 3\to 2$

$\frac{1}{\lambda }=109677.76\times 1^2\times [\frac{1}{3^2}-\frac{1}{4^2}]$

$\lambda =1875\times {10}^{-7}cm$.