Rydberg given the equation for all visible radiation in the hydrogen spectrum as $\lambda = \frac{kn

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Rydberg given the equation for all visible radiation in the hydrogen spectrum as $\lambda = \frac{kn^{2}}{n^{2} - 4}$. The value of k in terms of Rydberg constant is

Accepted Answer

$\frac{1}{\lambda} = R 	imes 1^{2}\left( \frac{1}{2^{2}} - \frac{1}{n^{2}} ight) = \frac{R.\left( n^{2} - 4 ight)}{4n^{2}}$$	herefore\lambda = \frac{4n^{2}}{R\left( n^{2} - 4 ight)} = \frac{K.n^{2}}{n^{2} - 4} \Rightarrow K = \frac{4}{R}$

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