Question

Hydrogen ${(}_1{H}^1),$ Deuterium ${(}_1{H}^2)$, singly ionised Helium ${(}_2{H}^4{)}^{+}$ and doubly ionised lithium ${(}_{3}L{i}^6{)}^{++}$ 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 ${\lambda }_1,{\lambda }_2,{\lambda }_3$ and ${\lambda }_4$ respectively then approximately which one of the following is correct ?

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

Answer 1

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

From Rydberg's formula, $\frac{1}{\lambda }=R{z}^2(\frac{1}{n_f^2}-\frac{1}{n_i^2})$

Given: $n_i=2,n_f=1$

1. For Hydrogen ${(}_1{H}^1)$
$\frac{1}{{\lambda }_1}=R(1{)}^2\left(\frac{3}{4}\right)...(1)$

2. Deuterium ${(}_1{H}^2)$,
$\frac{1}{{\lambda }_2}=R(1{)}^2\left(\frac{3}{4}\right)...(2)$

3. singly ionised Helium ${(}_2{H}^4{)}^{+}$
$\frac{1}{{\lambda }_3}=R(2{)}^2\left(\frac{3}{4}\right)...(3)$

4.doubly ionised lithium ${(}_{3}L{i}^6{)}^{++}$
$\frac{1}{{\lambda }_4}=R(3{)}^2\left(\frac{3}{4}\right)...(4)$

From eq. (1), (2), (3) and (4)
$\frac{1}{{\lambda }_1}=\frac{1}{{\lambda }_2}=\frac{1}{4{\lambda }_3}=\frac{1}{9{\lambda }_4}$
or ${\lambda }_1={\lambda }_2=4{\lambda }_3=9{\lambda }_4$

Hence, option $C$ is correct.