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(B) The Rydberg formula for the wavelength $\lambda$ of a spectral line is given by: $\frac{1}{\lambda} = R \left[ \frac{1}{n_2^2} - \frac{1}{n_1^2} \

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$\frac{100}{21 R}$

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(B) The Rydberg formula for the wavelength $\lambda$ of a spectral line is given by: $\frac{1}{\lambda} = R \left[ \frac{1}{n_2^2} - \frac{1}{n_1^2} \right]$.For the Balmer series, $n_2 = 2$ and $n_1 = 3, 4, 5, \dots$.Substituting $n_2 = 2$ and $n_1 = 5$ into the formula:$\frac{1}{\lambda} = R \left[ \frac{1}{2^2} - \frac{1}{5^2} \right] = R \left[ \frac{1}{4} - \frac{1}{25} \right]$.$\frac{1}{\lambda} = R \left[ \frac{25 - 4}{100} \right] = R \left[ \frac{21}{100} \right]$.Therefore, $\lambda = \frac{100}{21 R}$.

Which one of the following corresponds to the wavelength of a line in the Balmer series of the $H$ atom spectrum? $(R =$ Rydberg constant$)$

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