Question

Hydrogen $\left({}_1{H}^1\right)$, Deuterium $\left({}_1{H}^2\right)$, singly ionised Helium ${\left({}_{2}H{e}^4\right)}^{+}$ and doubly ionised lithium ${\left({}_{3}L{i}^6\right)}^{++}$ all have one electron around the nucleus. Consider an electron transition from $n=2$ to $n=1$. If the wave lengths of emitted radiation are ${\lambda }_1,{\lambda }_{2^{\prime }}{\lambda }_3$ and ${\lambda }_4$ respectively then approximately which one of the following is correct?

• A. $4{\lambda }_1=2{\lambda }_2,=2{\lambda }_3={\lambda }_4$
• B. ${\lambda }_1=2{\lambda }_2,=2{\lambda }_3={\lambda }_4$
• C. ${\lambda }_1={\lambda }_2,=4{\lambda }_3=9{\lambda }_4$
• D. ${\lambda }_1=2{\lambda }_2,=3{\lambda }_3=4{\lambda }_4$

Answer 1

Answer: (C)

$\frac{1}{\lambda }=R{Z}^2\left[\frac{1}{n_1^2}-\frac{1}{n^2}\right]$
$\Rightarrow \lambda \propto \frac{1}{Z^2}\text{ for given }{n}_1$ & $n_2$
$\Rightarrow {\lambda }_1={\lambda }_2=4{\lambda }_3=9{\lambda }_4$