Question

If the shortest wavelength of the spectral line of $H$-atom in the Lyman series is $x$, then the longest wavelength of the line in Balmer series of $L{i}^{2+}$ is

• A. $9x$
• B. $\frac{x}{9}$
• C. $\frac{5x}{4}$
• D. $\frac{4x}{5}$

Answer 1

Answer: (D)

$\frac{1}{\lambda }=R_H{Z}^2\left(\frac{1}{n_1{}^2}-\frac{1}{n_2^2}\right)$
To calculate shortest wavelength take $n_2=\infty$ and longest wavelength take nearest value of $n_2$.
For $H$-atom, to calculate
$\frac{1}{{\lambda }_{\text{shortest }}},n_2=\infty ,Z=1,n_1=1$
$\therefore \frac{1}{x}=R_H$ (Lyman series)
For $\frac{1}{{\lambda }_{\text{longest }}}$ for $L{i}^{2+},Z=3,n_1=2,n_2=3$ (Balmer series)
$\frac{1}{{\lambda }_{\text{longest }}}=\frac{1}{x}\times 3^2\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=\frac{5}{4x}$
$\therefore {\lambda }_{\text{longest }}=\frac{4x}{5}$