Question

The shortest wavelength of transition in paschen series of $H{e}^{+}$ ion in nanometer (nm) is$\frac{X}{R_H}$ nm, then $X$ is

Answer 1

For Paschen $n_1=3$
For shortest wavelength $n_2=\infty$
$\therefore \frac{1}{{\lambda }_{min}}=4\times R_H[\frac{1}{3^2}-\frac{1}{{\infty }^2}]\Rightarrow \frac{1}{{\lambda }_{min}}=\frac{4}{9}{R}_H\Rightarrow {\lambda }_{min}=\frac{9}{4}\times \frac{1}{R_H}nm=\frac{2.25}{R_H}$