Question

Shortest wavelength present in paschen series of li2+ ion

Answer 1

Answer: $0.91\times 10^{-7}m$

Explanation:

$E=\frac{hc}{\lambda}$

$\lambda$ = Wavelength of radiation

E= energy

For wavelength to be minimum, energy would be maximum, i.e the electron will jump from n=3  level to infinite level for Paschen series.

Using Rydberg's Equation:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )\times Z^2$

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant  =$1.097\times 10^7m^{-1}$

$n_f$ = Higher energy level = $\infty$  

$n_i$= Lower energy level = 3 (Paschen series)

Z= atomic number = 3 (for lithium)

Putting the values, in above equation, we get

$\frac{1}{\lambda_{lyman}}=R_H\left(\frac{1}{1^2}-\frac{1}{\infty^2} \right )\times 4$

$\frac{1}{\lambda}=1.097\times 10^7\left(\frac{1}{3^2}-\frac{1}{\infty^2} \right )\times 3^2$

$\lambda=0.91\times 10^{-7}m$

Thus shortest wavelength present in paschen series of $Li^{2+}$ is $0.91\times 10^{-7}m$