Question

What is the maximum wavelength of line of Balmer series of hydrogen spectrum? $(R=1.09\times {10}^7{m}^{-1})$

AIIMS 2018

Answer 1

$\frac{1}{\lambda }=R_H\times Z^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$
For maximum wavelength in the Balmer series
$n_2=3;n_1=2$
$\frac{1}{\lambda }=1.09\times {10}^7\times 1^2(\frac{1}{2^2}-\frac{1}{3^2})$
$\frac{1}{\lambda }=1.09\times {10}^7(\frac{1}{4}-\frac{1}{9})$
$\lambda =6.60\times {10}^{-7}m=660nm$