Question

Hydrogen (H), deuterium (4), 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. $\lambda_{1} = \lambda_{2} = 2\lambda_{3} = 3\lambda_{4}$

Answer 1

Answer: (A)

$\lambda_{1} = \lambda_{2} = 4\lambda_{3} = 9\lambda_{4}$

Answer 2

Using $\Delta E \propto Z^{2}$ (∵ $n_{1}$ and $n_{2}$ are same)

⇒ $\frac{hc}{\lambda} \propto Z^{2}$ ⇒ $\lambda Z^{2} = \text{constant}$

⇒ $\lambda_{1}Z_{1}^{2} = \lambda_{2}Z_{2}^{2} = \lambda_{3}Z_{3}^{2} = \lambda_{4}Z^{4}$

⇒ $\lambda_{1} \times 1 = \lambda_{2} \times 1^{2} = \lambda_{3} \times 2^{2} = \lambda_{4} \times 3^{3}$

⇒ $\lambda_{1} = \lambda_{2} = 4\lambda_{3} = 9\lambda_{4}$.