Question

The wavenumber for the shortest wavelength transition in paschen series of Li2+ ion is

Answer 1

shortest wavelength transition in Paschen series of $Li^{2+}$ ion is given by, $\frac{1}{\lambda}=\frac{RZ^2}{3^2}$

where R = 1.097374 × 10^7 m^-1
Z is the atomic number of hydrogen like ion.

here Z = 3

so, Wavelength transition in Paschen series of Li2+ ion $\frac{1}{\lambda}=\frac{R(3)^2}{3^2}$
$\implies\frac{1}{\lambda}=R$

we know, wavenumber is the inverse of wavelength. e.g., $\overline{\nu}=\frac{1}{\lambda}$

so, wavenumber = R = 1.097374 × 10^7 m^-1