Question

Hydrogen $(H)$, deuterium $(D)$, singly ionized helium $\left(H{e}^{+}\right)$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. ${\lambda }_1={\lambda }_2=2{\lambda }_3=3\sqrt{2}{\lambda }_4$

Answer 1

Answer: (A)

Using $\Delta E\propto Z^2$
$(\therefore n_1$ and $n_2$ are same)
$\Rightarrow \frac{hc}{\lambda }\propto Z^2$
$\Rightarrow \lambda Z^2=$ constant
$\Rightarrow {\lambda }_1{Z}_1^2={\lambda }_2{Z}_2^2={\lambda }_3{Z}_3^2={\lambda }_4{Z}_4^2$
$\Rightarrow {\lambda }_1\times 1={\lambda }_2\times 1^2={\lambda }_3\times 2^2={\lambda }_3\times 3^3$
$\Rightarrow {\lambda }_1={\lambda }_{2}4{\lambda }_3=9{\lambda }_4$