In the hydrogen spectrum,the ratio of the wav | vedclass

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(C) The wavelength $\lambda$ for a transition in the hydrogen spectrum is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2}

Vedclass · Accepted Answer

$0.25$

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(C) The wavelength $\lambda$ for a transition in the hydrogen spectrum is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$,where $R$ is the Rydberg constant.For the last line of the Lyman series,$n_1 = 1$ and $n_2 = \infty$. Thus,$\frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R$,which gives $\lambda_L = \frac{1}{R}$.For the last line of the Balmer series,$n_1 = 2$ and $n_2 = \infty$. Thus,$\frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = \frac{R}{4}$,which gives $\lambda_B = \frac{4}{R}$.The ratio of the wavelengths is $\frac{\lambda_L}{\lambda_B} = \frac{1/R}{4/R} = \frac{1}{4} = 0.25$.

In the hydrogen spectrum,the ratio of the wavelengths of the last line of the Lyman series and that of the last line of the Balmer series is:

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