Question

The wave number for the shortest wavelength transition in Paschal series of ${Li}^{2+}$ Ion is: (R: 109677 ${cm}^{-1}$)

• A. $27420$ ${cm}^{-1}$
• B. $164515$ ${cm}^{-1}$
• C. $219354$ ${cm}^{-1}$
• D. $109677$ ${cm}^{-1}$

Answer 1

Answer: (D)

$109677$ ${cm}^{-1}$

Given that the transition takes place in the Paschen series.

$\Rightarrow n_1=3$

The wavelength of the emitted photon is shortest.

$\Rightarrow n_2=\infty$ (i.e. energy is maximum)

Rydberg constant for $H,R_H=109677c{m}^{-1}$

Rydberg constant for $L{i}^{2+},$

$R_{L{i}^{2+}}=R_H\times Z$

Where, $Z=$ atomic number of $Li$

Now,

$\overline{\upsilon }=\frac{1}{\lambda }=R_H\times Z^2[\frac{1}{n_{f^2}}-\frac{1}{n_{i^2}}]$

$=109677c{m}^{-1}\times 3^2[\frac{1}{3^2}-\frac{1}{{\infty }^2}]$

$=109677c{m}^{-1}\times 3^2\times [\frac{1}{3^2}-0]$

$=109677c{m}^{-1}$

Hence, option D is correct.