Chemistry - Atomic Structure Question with Solution | TestHub

For hydrogen-like species,

$$\frac{1}{\lambda }=R{Z}^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$

For He⁺ (Z=2), transition $2\to 4$:

$$\frac{1}{\lambda }=R(2{)}^2\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=4R\left(\frac{1}{4}-\frac{1}{16}\right)=4R\left(\frac{3}{16}\right)=\frac{3R}{4}$$

For Li²⁺ (Z=3), let the required transition be $n\to 2n$ (check option C: $3\to 6$):

$$\frac{1}{\lambda }=R(3{)}^2\left(\frac{1}{3^2}-\frac{1}{6^2}\right)=9R\left(\frac{1}{9}-\frac{1}{36}\right)=9R\left(\frac{3}{36}\right)=\frac{3R}{4}$$

The wavelengths are identical.

Answer: (C) $3\to 6$