Question

Find the shortest wavelength in Paschen series if, the longest wavelength in Balmer series is 6563 $A^0$

Answer 1

Given: Longest wavelength of Balmer series $=6563A^o$

To find: Shortest wavelength of Paschen series $=?$

Soln:

We know that

$\frac{1}{\lambda }=R{Z}^2(\frac{1}{n^2}-\frac{1}{m^2})$

For Balmer series

$\frac{1}{\lambda }=R{Z}^2(\frac{1}{2^2}-\frac{1}{3^2})\Rightarrow \frac{1}{6563}=R{Z}^2(\frac{1}{4}-\frac{1}{9})$

$\Rightarrow \frac{1}{6563}=R{Z}^2\left(\frac{9-4}{36}\right)$

$\Rightarrow R{Z}^2=\frac{36}{5\times 6563}$ ...(1)

For paschen series

Shortest wavelength when $(m=\infty )$

$\frac{1}{{\lambda }_{min}}=R{Z}^2(\frac{1}{3^2}-\frac{1}{\infty })$

$\frac{1}{{\lambda }_{min}}=\frac{36}{5\times 6563}\times \frac{1}{9}$

${\lambda }_{min}=8203.75A^o$