Question

The shortest wavelength for Lyman series is $912\stackrel{o}{A}$. Find shortest wavelength for Paschen and Brackett series in Hydrogen atom.

Answer 1

Given shortest wavelength $\lambda ={912}^{\circ }A$

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

For shortest wavelength in Paschen series the two levels showing transition should be maximum. $n=\infty$

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

$\lambda =8.20\times {10}^{-7}m=8200A^{\circ }$

For shortest wavelength in Brackett series the two levels showing transition should be maximum. $n=\infty$

$\frac{1}{\lambda }=1.097\times {10}^7(\frac{1}{4^2}-\frac{1}{{\infty }^2})$

$\lambda =14.58\times {10}^{-7}m=14580A^{\circ }$