The wavelength of the first line in the Balme | vedclass

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(A) The Rydberg formula for the hydrogen spectrum is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$,where $R$ is th

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$\frac{20\lambda}{27}$

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(A) The Rydberg formula for the hydrogen spectrum is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$,where $R$ is the Rydberg constant.For the Balmer series,$n_1 = 2$. The first line corresponds to the transition from $n_2 = 3$ to $n_1 = 2$.$\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = R \left( \frac{5}{36} ight) \Rightarrow \lambda = \frac{36}{5R}$.The second line corresponds to the transition from $n_2 = 4$ to $n_1 = 2$.$\frac{1}{\lambda'} = R \left( \frac{1}{2^2} - \frac{1}{4^2} ight) = R \left( \frac{1}{4} - \frac{1}{16} ight) = R \left( \frac{3}{16} ight) \Rightarrow \lambda' = \frac{16}{3R}$.Dividing $\lambda'$ by $\lambda$:$\frac{\lambda'}{\lambda} = \frac{16}{3R} 	imes \frac{5R}{36} = \frac{16 	imes 5}{3 	imes 36} = \frac{80}{108} = \frac{20}{27}$.Therefore,$\lambda' = \frac{20}{27}\lambda$.

The wavelength of the first line in the Balmer series in the hydrogen spectrum is $\lambda$. What is the wavelength of the second line?

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