Question

The shortest wavelength of He atom in Balmer series is $x$, then longest wavelength in the Paschen series of $L{i}^{+2}$ is:

• A. $\frac{36x}{5}$
• B. $\frac{16x}{7}$
• C. $\frac{9x}{5}$
• D. $\frac{5x}{9}$

Answer 1

The correct option is B $\frac{16x}{7}$

The wavelength of $H$-like an atom is given by

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

For shortest wavelength $x$ of Paschen series $n_2=\infty$ and $n_1=3$

$\frac{1}{x}=R{\left(3\right)}^2[\frac{1}{3^2}-\frac{1}{{\infty }^2}]$

$\Rightarrow x=\frac{9}{9R}=\frac{1}{R}$

For longest wavelength of Paschen series $n_2=4$ and $n_1=3$

$\frac{1}{\lambda }=R{\left(3\right)}^2[\frac{1}{3^2}-\frac{1}{4^2}]$

$\Rightarrow \frac{1}{\lambda }=R{\left(3\right)}^2\left[\frac{(16-9)}{16\times 9}\right]$

$\Rightarrow \lambda =\frac{16}{7R}=\frac{16x}{7}$

Hence, the correct option is $B$