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(A) The wavelength for the Lyman series is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{n^2} \right)$.For the

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$\frac{27}{5} \lambda$

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(A) The wavelength for the Lyman series is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{n^2} \right)$.For the first line of the Lyman series,the transition is from $n = 2$ to $n = 1$.Substituting these values: $\frac{1}{\lambda} = R \left( \frac{1}{1} - \frac{1}{4} \right) = R \left( \frac{3}{4} \right)$.Thus,$R = \frac{4}{3\lambda}$ (Equation $i$).The wavelength for the Balmer series is given by: $\frac{1}{\lambda^{\prime}} = R \left( \frac{1}{2^2} - \frac{1}{n^2} \right)$.For the first line of the Balmer series,the transition is from $n = 3$ to $n = 2$.Substituting these values: $\frac{1}{\lambda^{\prime}} = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{9 - 4}{36} \right) = R \left( \frac{5}{36} \right)$.Substituting the value of $R$ from Equation $i$: $\frac{1}{\lambda^{\prime}} = \left( \frac{4}{3\lambda} \right) \left( \frac{5}{36} \right) = \frac{20}{108\lambda} = \frac{5}{27\lambda}$.Therefore,$\lambda^{\prime} = \frac{27}{5} \lambda$.

List $I$ (Spectral Lines of Hydrogen for transitions from)	List $II$ (Wavelengths $(nm)$)
$A$. $n_2=3$ to $n_1=2$	$I$. $410.2$
$B$. $n_2=4$ to $n_1=2$	$II$. $434.1$
$C$. $n_2=5$ to $n_1=2$	$III$. $656.3$
$D$. $n_2=6$ to $n_1=2$	$IV$. $486.1$

If the first line in the Lyman series has wavelength $\lambda$,then the first line in the Balmer series has the wavelength

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