Question

Hydrogen $(H)$, deuterium $(D)$, singly ionized helium $\left(H e^{+}\right)$and doubly ionized lithium $(L i)$ 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. $\lambda_{1}=\lambda_{2}=2 \lambda_{3}=3 \sqrt{2} \lambda_{4}$

Answer 1

Answer: (A)

Using $\Delta E \propto Z^{2}$
$\left(\therefore n_{1}\right.$ and $n_{2}$ are same)
$\Rightarrow \frac{h c}{\lambda} \propto Z^{2}$
$ \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_{3} \times 3^{3} $
$\Rightarrow \lambda_{1}=\lambda_{2} 4 \lambda_{3}=9 \lambda_{4}$