If the shortest wavelength of the He^{+} ion | vedclass

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(B) The Rydberg formula is given by $\frac{1}{\lambda} = R Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$.For the shortest wavelength in the Balm

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$\frac{16}{7} \ X$

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(B) The Rydberg formula is given by $\frac{1}{\lambda} = R Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}ight)$.For the shortest wavelength in the Balmer series of $He^{+}$ $(Z=2)$: $n_1 = 2$,$n_2 = \infty$.$\frac{1}{X} = R 	imes 2^2 \left(\frac{1}{2^2} - \frac{1}{\infty^2}ight) = R 	imes 4 	imes \frac{1}{4} = R$.So,$R = \frac{1}{X}$.For the longest wavelength in the Paschen series of $Li^{2+}$ $(Z=3)$: $n_1 = 3$,$n_2 = 4$.$\frac{1}{\lambda_{Li^{2+}}} = R 	imes 3^2 \left(\frac{1}{3^2} - \frac{1}{4^2}ight) = R 	imes 9 \left(\frac{1}{9} - \frac{1}{16}ight) = R 	imes 9 \left(\frac{16-9}{144}ight) = R 	imes 9 	imes \frac{7}{144} = R 	imes \frac{7}{16}$.Substituting $R = \frac{1}{X}$:$\frac{1}{\lambda_{Li^{2+}}} = \frac{1}{X} 	imes \frac{7}{16} \implies \lambda_{Li^{2+}} = \frac{16}{7} \ X$.

If the shortest wavelength of the $He^{+}$ ion in the Balmer series is $X \ m$,then the longest wavelength in the Paschen series of the $Li^{2+}$ ion is:

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