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(B) The Rydberg formula is given by $\frac{1}{\lambda} = R_H 	imes Z^2 	imes (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.For the longest wavelength of the

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$\frac{7}{12} x$

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(B) The Rydberg formula is given by $\frac{1}{\lambda} = R_H 	imes Z^2 	imes (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.For the longest wavelength of the Paschen series of $Li^{2+}$ $(Z=3)$,we have $n_1 = 3$ and $n_2 = 4$.$\frac{1}{x} = R_H 	imes (3)^2 	imes (\frac{1}{3^2} - \frac{1}{4^2}) = R_H 	imes 9 	imes (\frac{1}{9} - \frac{1}{16}) = R_H 	imes 9 	imes \frac{7}{144} = R_H 	imes \frac{7}{16}$.Thus,$x = \frac{16}{7 R_H}$ ...$(i)$.For the longest wavelength of the Lyman series of the hydrogen spectrum $(Z=1)$,we have $n_1 = 1$ and $n_2 = 2$.$\frac{1}{\lambda} = R_H 	imes (1)^2 	imes (\frac{1}{1^2} - \frac{1}{2^2}) = R_H 	imes (1 - \frac{1}{4}) = R_H 	imes \frac{3}{4}$.Thus,$\lambda = \frac{4}{3 R_H}$ ...$(ii)$.Dividing equation $(ii)$ by equation $(i)$:$\frac{\lambda}{x} = \frac{4 / (3 R_H)}{16 / (7 R_H)} = \frac{4}{3} 	imes \frac{7}{16} = \frac{7}{12}$.Therefore,$\lambda = \frac{7}{12} x$.

Energy level	$K$ $(n=1)$	$L$ $(n=2)$	$M$ $(n=3)$	$N$ $(n=4...n=\infty)$
Energy	$-864 \ a.u.$	$-216 \ a.u.$	$-96 \ a.u.$	$0 \ a.u.$

If the longest wavelength of the spectral line of the Paschen series of $Li^{2+}$ ion spectrum is $x \ \mathring{A}$,then the longest wavelength (in $\mathring{A}$) of the Lyman series of the hydrogen spectrum is

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