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The first emission line of Balmer series for H – spectrum has the wave number equal to

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$\frac{7 R_{H}}{144} c m^{- 1}$

$\frac{3 R_{H}}{4} c m^{- 1}$

$\frac{9 R_{H}}{400} c m^{- 1}$

$\frac{5 R_{H}}{36} c m^{- 1}$

answer is D.

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Detailed Solution

$\bar{v} = \frac{1}{λ} = R_{H} [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}] R_{H} = 109677 c m^{- 1}$

$\begin{array}{l} n_{1} = 2, n_{2} = 3 \\ \bar{v} = R_{H} [\frac{1}{2^{2}} - \frac{1}{3^{2}}] \\ \bar{v} = R_{H} [\frac{1}{4} - \frac{1}{9}] \\ \bar{v} = R_{H} [\frac{9 - 4}{36}] \\ \bar{v} = \frac{5 R_{H}}{36} c m^{- 1} \end{array}$

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