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Q. Assertion : Balmer series lies in the visible region of electromagnetic spectrum.
Reason : $\frac{1}{\lambda} = R (\frac {1}{2^2}-\frac {1}{n^2})$

AIIMSAIIMS 2012

Solution:

The wavelength in Balmer series is given by
$\frac{1}{\lambda} =R\left(\frac{1}{2^{2}}-\frac{1}{n^{2}}\right), n=3,4,5 $
$\frac{1}{\lambda_{\max }} =R\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right) $
$\frac{1}{\lambda_{\max }} =\frac{36}{5 R}$
$=\frac{36}{5 \times 1.097 \times 10^{7}}=6563 \,\mathring{A}$
and $ \frac{1}{\lambda_{\min }}=R\left(\frac{1}{2^{2}}-\frac{1}{\infty^{2}}\right)$
$\frac{1}{\lambda_{\min }}=\frac{4}{R}=\frac{4}{1.097 \times 10^{7}}=3646\, \mathring{A}$
The wavelength $6563 \,\mathring{A}$ and $3646 \,\mathring{A}$ lie in visible region.
Therefore, Balmer series lies in visible region.