Question

Hydrogen ${(}_1{H}^1)$, Deuterium ${(}_1{H}^2)$, singly ionised Helium ${(}_{2}H{e}^4{)}^{+}$ and double 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. $\lambda ={\lambda }_2=4{\lambda }_3=9{\lambda }_4$
• B. ${\lambda }_1=2{\lambda }_2=3{\lambda }_3=4{\lambda }_4$
• C. $4{\lambda }_1=2{\lambda }_2=2{\lambda }_3={\lambda }_4$
• D. ${\lambda }_1=2{\lambda }_2=2{\lambda }_3={\lambda }_4$

Answer 1

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

Answer is A
We know that,
$\frac{1}{\lambda }=R{z}^2(1/1^2-1/2^2)$
z= 1 for hydrogen and deuterium
z=2 for singly ionised Helium
z=3 for double ionised lithium
Soving above, we get
${\lambda }_1={\lambda }_2=4{\lambda }_3=9{\lambda }_4$