The ratio of the wavelengths of the first Lym | vedclass

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

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$1: 4$

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(B) The wavelength $\lambda$ for a transition in a hydrogen atom 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.$1$. For the first Lyman line,the transition is from $n_2 = 2$ to $n_1 = 1$:$\frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = \frac{3R}{4} \implies \lambda_L = \frac{4}{3R}$.$2$. For the second Balmer line,the transition is from $n_2 = 4$ to $n_1 = 2$:$\frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{4} - \frac{1}{16} \right) = R \left( \frac{4-1}{16} \right) = \frac{3R}{16} \implies \lambda_B = \frac{16}{3R}$.$3$. The ratio of the wavelengths is:$\frac{\lambda_L}{\lambda_B} = \frac{4/3R}{16/3R} = \frac{4}{16} = \frac{1}{4}$.Thus,the ratio is $1: 4$.

The ratio of the wavelengths of the first Lyman line and the second Balmer line of the hydrogen atom is

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