Question

What is the shortest wavelength line in the paschen series of li2+?

Answer 1

see there is ur answer

Answer 2

The shortest wavelength in the given paschen lines is 820 nm

SOLUTION:

The wavelength in the Paschen series is known by the transition state and can be found using the relation.

$\frac{1}{\lambda}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$

Where n is the transition state and R is the Rydbergs constant.

For lithinium ion having +2 electrons the transition state will be as  

$n_{1}=3\ And\ n_{2}=\infty$

Substituting the values in the relation,

$\frac{1}{\lambda}=R\left(\frac{1}{3^{2}}-\frac{1}{\infty^{2}}\right)$

The value for Rydberg constant is $1.09 \times 10^{7} m$

We get the wavelength corresponding to

$\lambda=\frac{9}{1.09 \times 10^{7}}=8.20 \times 10^{-7} \mathrm{m}=820 \mathrm{nm}$