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(C) The Rydberg formula for a hydrogen-like ion is given by: $1 / \lambda = R Z^{2} [ 1 / n_{1}^{2} - 1 / n_{2}^{2} ]$For the Paschen series,the trans

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$1 / R$

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(C) The Rydberg formula for a hydrogen-like ion is given by: $1 / \lambda = R Z^{2} [ 1 / n_{1}^{2} - 1 / n_{2}^{2} ]$For the Paschen series,the transition ends at $n_{1} = 3$.For the shortest wavelength,the transition starts from $n_{2} = \infty$.For $Li^{2+}$ ion,the atomic number $Z = 3$.Substituting these values: $1 / \lambda = R 	imes 3^{2} [ 1 / 3^{2} - 1 / \infty^{2} ]$$1 / \lambda = R 	imes 9 [ 1 / 9 - 0 ]$$1 / \lambda = R 	imes 9 	imes (1 / 9) = R$Therefore,$\lambda = 1 / R$.

What is the shortest wavelength for the Paschen series of the $Li^{2+}$ ion?

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