Question

The maximum wavelength of Brakett series of hydrogen atom will be ( in Angstroms)

Answer 1

For Brackett series, $n_1=4$ and $n_2=5.6,7,.........,\infty$
Hence, the wavelengths of Brackett series are given by the formula:
$\frac{1}{\lambda }=R[\frac{1}{4^2}-\frac{1}{n_2^2}]$

For maximum wavelength i.e. for the first member of the series, $n_2=5$
Therefore,
$\frac{1}{\lambda }=R[\frac{1}{4^2}-\frac{1}{5^2}]$

$\frac{1}{\lambda }=R[\frac{1}{16}-\frac{1}{25}]$

$\frac{1}{\lambda }=R\left[\frac{9}{400}\right]$

$\lambda =\frac{400}{9R}$

$\lambda =\frac{400}{9\times 10.97\times {10}^6}$

$\lambda =\frac{400}{98.73}\times {10}^{-6}$

$\lambda =4.0514\times {10}^{-6}$

$\lambda =40514\times {10}^{-10}$

$\lambda =40514\mathring{A}$