Question

Find $\lambda$ of first line of balmer series

Answer 1

$\lambda =654.5nm$

Wave number of emitted radiation is given by

$\overline{v}=\frac{1}{\lambda }=R_H{Z}^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

For Balmer series,

$n_1=r$

ans for $I^{st}$ member of Balmer series. i.e.,

$H_{\alpha },n_2=3$

$\therefore \overline{v}=\frac{1}{\lambda }=1.1\times {10}^7\times (1{)}^2[\frac{1}{4}-\frac{1}{9}]$

or $\frac{1}{\lambda }=\frac{1.1\times {10}^7}{36}\times 5$

$\Rightarrow \lambda =\frac{36\times {10}^{-7}}{5\times 1.1}=654.5\times {10}^{-9}$