1)

Hydrogen (1H1), deuterium (1H2), singly ionised helium   $(_{2}He^{4})^{+}$   and doubly ionised lithium$(_{3}Li^{8})^{++}$ all have one electron around the nucleus. Consider an electron transition from n=2 to n=1 . If the wavelength s of emitted radiation are   $\lambda_{1},\lambda_{2},\lambda_{3}$ and   $\lambda_{4}$, respectively for four elements, 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) $\lambda_{1}=2\lambda_{2}=3\lambda_{3}=4\lambda_{4}$

Option C

### Explanation:

For hydrogen atom, we get

$\frac{1}{\lambda}=Rz^{2}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)\Rightarrow\frac{1}{\lambda_{1}}=R(1)^{2}\left(\frac{3}{4}\right)$

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

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

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

$\Rightarrow\frac{1}{\lambda_{1}}=\frac{1}{4\lambda_{3}}=\frac{1}{9\lambda_{4}}=\frac{1}{\lambda_{2}}$