Balmer gives an equation for wavelength of visible radiation of H-spectrum as $λ = \frac{{kn}^{2}}{(n^{2} - 4)} .$ The value of k in terms of Rydberg’s constant R is

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R/4

4/R

answer is D.

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Detailed Solution

$\frac{1}{λ} = R Z^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) . For H - atom Z = 1 .$

For visible radiation, $n_{1} = 2$

$∴ \frac{1}{λ} = R (\frac{1}{2_{}^{2}} - \frac{1}{n_{}^{2}}) \Rightarrow λ = \frac{4 n^{2}}{R (n^{2} - 4)} = \frac{k n^{2}}{(n^{2} - 4)} \Rightarrow k = \frac{4}{R} .$

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