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(B) The wavelength of visible radiation in the Balmer series is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{m

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$\frac{4}{R}$

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(B) The wavelength of visible radiation in the Balmer series is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{m^2} \right)$ for $m = 3, 4, 5, ...$ $\frac{1}{\lambda} = R \left( \frac{1}{4} - \frac{1}{m^2} \right)$ $\frac{1}{\lambda} = R \left( \frac{m^2 - 4}{4 m^2} \right)$ Taking the reciprocal to find $\lambda$: $\lambda = \frac{4 m^2}{R(m^2 - 4)}$ Comparing this with the given equation $\lambda = \frac{k m^2}{m^2 - 4}$,we get: $k = \frac{4}{R}$

For the wavelength of visible radiation of the hydrogen spectrum,Balmer gave an equation as $\lambda = \frac{(k m^2)}{(m^2 - 4)}$,where $m$ is an integer. The value of $k$ in terms of Rydberg's constant $R$ is

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