Question

If the shortest wavelength in Lyman series of hydrogen atom is A, then the longest wavelength in Paschen series of $H{e}^{+}$ is :

• A. $\frac{36A}{7}$
• B. $\frac{5A}{9}$
• C. $\frac{36A}{5}$
• D. $\frac{9A}{5}$

Answer 1

The correct option is A $\frac{36A}{7}$

For Lyman series $hydrogen$ atom, the shortest wavelength corresponds to:

$n_1=1$ and $n_2=\infty$

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

$\frac{1}{A}=R[\frac{1}{1^2}-\frac{1}{{\infty }^2}]$

$\frac{1}{A}=R$ .....(1)

For Paschen series of $H{e}^{+}$ ion, the longest wavelength corresponds to :

$n_1=3$ and $n_2=4$

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

$\frac{1}{\lambda }=\frac{1}{A}\times 2^2\times [\frac{1}{3^2}-\frac{1}{4^2}]$

$\lambda =\frac{36A}{7}$