The wavelength of the first line in the Balme | vedclass

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(A) The Rydberg formula for the Balmer series is given by $\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{n^2} \right)$,where $n = 3, 4, 5, \do

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

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(A) The Rydberg formula for the Balmer series is given by $\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{n^2} ight)$,where $n = 3, 4, 5, \dots$.For the first line of the Balmer series,$n = 3$:$\frac{1}{\lambda} = R \left( \frac{1}{4} - \frac{1}{9} ight) = R \left( \frac{9-4}{36} ight) = \frac{5R}{36} \implies R = \frac{36}{5\lambda}$.For the second line of the Balmer series,$n = 4$:$\frac{1}{\lambda'} = R \left( \frac{1}{4} - \frac{1}{16} ight) = R \left( \frac{4-1}{16} ight) = \frac{3R}{16}$.Substituting the value of $R$ from the first equation into the second:$\frac{1}{\lambda'} = \frac{3}{16} 	imes \left( \frac{36}{5\lambda} ight) = \frac{3 	imes 9}{4 	imes 5 \lambda} = \frac{27}{20\lambda}$.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 in the same series?

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