Question

The minimum wavelength of Paschen series of hydrogen atom will be

• A. $18,700A^{\circ }$
• B. $970A^{\circ }$
• C. $1,022A^{\circ }$
• D. $8,181A^{\circ }$

Answer 1

Answer: (D)

$8,181A^{\circ }$

For Paschen series, $n_1=3$ and $n_2=4,5.6,7,.........,\infty$
Hence, the wavelengths of Paschen series are given by the formula:
$\frac{1}{\lambda }=R[\frac{1}{3^2}-\frac{1}{n_2^2}]$
For minimum wavelength i.e. for the series limit, $n_2=\infty$
Therefore,
$\frac{1}{\lambda }=R[\frac{1}{3^2}-\frac{1}{{\infty }^2}]$
$\frac{1}{\lambda }=R\left[\frac{1}{9}\right]$
$\lambda =\frac{9}{R}$
$\lambda =\frac{9}{11\times {10}^6}$
$\lambda =0.8181\times {10}^{-6}$
$\lambda =8181A^{\circ }$