Question

Which transition in $L{i}^{2+}$ would have the same wavelength as the $2\to 4$ transition in $H{e}^{+}$ ion?

• A. 4 to 6
• B. 4 to 3
• C. 5 to 6
• D. 5 to 4

Answer 1

Answer: (A)

4 to 6

Wavelength Formula

$\frac{1}{\lambda }=R_H{Z}^2(\frac{1}{n_i^2}-\frac{1}{n_f^2})$

given

${\lambda }_{l{i}^{+2}}={\lambda }_{H{e}^{+}}(2\to 4transition)$

Wavelength of $H{e}^{+}$ ion $(z=2)$

$\frac{1}{\lambda }=R_H\times 4(\frac{1}{4}-\frac{1}{16})$

$\frac{1}{\lambda }=4R_H\frac{3}{16}$

$\frac{3R_H}{4}$

Wavelength of $L{i}^{+2}$ ion $(Z=3)$

$\frac{1}{\lambda }=R_H\times 9(\frac{1}{n{i}^2}-\frac{1}{n{f}^2})$

$\frac{3R_H}{4}=R_H\times 9(\frac{1}{n{i}^2}-\frac{1}{n{f}^2})$

$\frac{1}{n{i}^2}-\frac{1}{n{f}^2}=\frac{1}{12}$