Question

Hydrogen (H), deuterium (D), singly ionized helium $\left(H{e}^{+}\right)$ and doubly ionized lithium $\left(L{i}^{++}\right)$ 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

The correct option is A ${\lambda }_1={\lambda }_2=4{\lambda }_3=9{\lambda }_4$

Using $\Delta E\propto Z^2$ ($\because n_{1}and{n}_2$ are same)
$\Rightarrow \frac{hc}{\lambda }\infty Z^2\Rightarrow \lambda Z^2=constant\Rightarrow {\lambda }_1{Z}_1^2={\lambda }_2{Z}_2^2={\lambda }_3{Z}_3^2={\lambda }_4{Z}^4\Rightarrow {\lambda }_1\times 1={\lambda }_2\times 1^2={\lambda }_3\times 2^2={\lambda }_4\times 3^3$
$\Rightarrow {\lambda }_1={\lambda }_2=4{\lambda }_3=9{\lambda }_4$