Question

Hydrogen $\left({}_1^{1}H\right)$, deuterium $\left({}_1^{2}H\right)$, singly ionised helium $\left({}_2^4{He}^{+}\right)$ and doubly ionised lithium $\left({}_3^6{Li}^{2+}\right)$ all have one electron around the nucleus. Consider an electron transition from $n=2$ to $n=1$. If the wavelength of emitted radiation are ${\lambda }_1,{\lambda }_2,{\lambda }_3$ and ${\lambda }_4$ respectively. Then approximately which one of the following is correct?

• A. ${\lambda }_1={\lambda }_2=4{\lambda }_3=9{\lambda }_4$
• 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 ${\lambda }_1={\lambda }_2=4{\lambda }_3=9{\lambda }_4$

For hydrogen atom,

we know

$\frac{1}{\lambda }={RZ}^2(\frac{1}{n_L^2}-\frac{1}{n_Z^2})$

We get

$\frac{1}{\lambda }={RZ}^2(\frac{1}{1^2}-\frac{1}{2^2})$

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

$\frac{1}{{\lambda }_2}=R{\left(1\right)}^2\left(\frac{3}{4}\right)$

$\frac{1}{{\lambda }_3}=R{\left(2\right)}^2\left(\frac{3}{4}\right)$

$\frac{1}{{\lambda }_4}=R{\left(3\right)}^2\left(\frac{3}{4}\right)$

$\frac{1}{{\lambda }_1}=\frac{1}{{4\lambda }_3}=\frac{1}{{9\lambda }_4}=\frac{1}{{\lambda }_2}$

$\Rightarrow {\lambda }_1={\lambda }_2={4\lambda }_3={9\lambda }_4$