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(B) The Rydberg formula for the Hydrogen spectrum is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.For the Balmer

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$\frac{4}{R}$

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(B) The Rydberg formula for the Hydrogen spectrum is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.For the Balmer series, $n_1 = 2$ and $n_2 = m$, where $m = 3, 4, 5, \dots$.Substituting these values, we get $\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{m^2} \right) = R \left( \frac{1}{4} - \frac{1}{m^2} \right)$.Simplifying the expression, $\frac{1}{\lambda} = R \left( \frac{m^2 - 4}{4m^2} \right)$.Taking the reciprocal, $\lambda = \frac{4m^2}{R(m^2 - 4)}$.Comparing this with the given equation $\lambda = \frac{xm^2}{m^2 - 4}$, we find that $x = \frac{4}{R}$.

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

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