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# Calculate the wave number for the shortest wavelength transition in the Balmer series of atomic hydrogen .

$\begin{array}{1 1} (a)\;27419.5\;cm^{-1} \\ (b)\;17419.5\;cm^{-1} \\ (c)\;7419.5\;cm^{-1}\\(d)\;27419.5\;cm^{-1} \end{array}$

The shortest wavelength transition corresponds to $n_2 \infty$ to $n_1=2$ transition .
$\bar{v}=109678 \bigg[ \large\frac{1}{n_1^2}-\frac{1}{n_2^2}\bigg]$$cm^{-1}$
$\qquad= 109678 \bigg[\large\frac{1}{(2)^2}-\frac{1}{\infty}\bigg]$
$\qquad= \large\frac{109678}{4}$
$\qquad= 27419.5\;cm^{-1}$