Question

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

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

Answer 1

The correct option is D $\frac{4y}{5}$

$\frac{1}{\lambda }=R{Z}^2(\frac{1}{n^2}-\frac{1}{m^2})$

For shortest lymen wavelength in H,

$m=\infty ,n=1$

$\frac{1}{\lambda }=R$

$R=\frac{1}{y}$

For longest balmer wavelength in $L{i}^{+2}$

$m=3,n=2$

$\frac{1}{\lambda }=R{3}^2(\frac{1}{4}-\frac{1}{9})=\frac{5R}{4}=\frac{5}{4y}$

$\lambda =\frac{4y}{5}$

Option $\text{D}$ is the correct answer.