In hydrogen spectrum the ratio of wavelengths of the last line of...

The wavelength of a spectral line in the hydrogen spectrum is given by Rydberg's formula:

$\dfrac{1}{\lambda} = R$\left( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right)

For$the last line of the Lyman series, the transition is from n_2 =$\infty$to n_1 = 1$.

\dfrac{1}{\lambda_L} =$R \left( \dfrac{1}{1^2} - \dfrac{1}{\infty^2} \right) =$ R$<br><br>$\lambda_L = $\dfrac{1}{R}<br><br>For$ the last line of the Balmer series, the transition is from n_2 = $\infty$ to n_1 = 2$.<br><br>\dfrac{1}{\lambda_B} =$ R \left( \dfrac{1}{2^2} - \dfrac{1}{\infty^2} \right) = \dfrac{R}{4}

$\lambda_B =$\dfrac{4}{R}

The ratio$of$the wavelengths is:

\dfrac{\lambda_L}{\lambda_B} = \dfrac{\dfrac{1}{R}}{\dfrac{4}{R}} = \dfrac{1}{4} =0.25$<br><br>Answer:$0.25