Q. Hydrogen $(H)$, deuterium $(D)$, singly ionized helium $(He^{+})$ and doubly ionized lithium $(Li^{+ +})$ all have one electron around the nucleus. Consider $n = 2$ to $n = 1$ transition. The wavelengths of emitted radiations are $\lambda_{1},\lambda_{2},\lambda_{3}$ and $\lambda_{4}$ respectively. Then approximately
A) $\lambda_{1} = \lambda_{2} = 4\lambda_{3} = 9\lambda_{4}$
B) $4\lambda_{1} = 2\lambda_{2} = 2\lambda_{3} = \lambda_{4}$
C) $\lambda_{1} = 2\lambda_{2} = 2\sqrt{2}\lambda_{3} = 3\sqrt{2}\lambda_{4}$
D) $\lambda_{1} = \lambda_{2} = 2\lambda_{3} = 3\sqrt{2}\lambda_{4}$
Explanation
(a) Using $\mathrm{\Delta}E \propto Z^{2}$ [$\therefore n_{1}$ and $n_{2}$ are same]
$\Rightarrow \frac{hc}{\lambda} \propto Z^{2} \Rightarrow \lambda Z^{2}$ = constant
$\Rightarrow \lambda_{1}Z_{1}^{2} = \lambda_{2}Z_{2}^{2} = \lambda_{3}Z_{3}^{2} = \lambda_{4}Z_{4}^{2}$
$\Rightarrow \lambda_{1} \times 1 = \lambda_{2} \times 1^{2} = \lambda_{3} \times 2^{2} = \lambda_{4} \times 3^{3}$
$\Rightarrow \lambda_{1} = \lambda_{2} = 4\lambda_{3} = 9\lambda_{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.