Physics - Modern Physics Question with Solution | TestHub

From Bohr’s formula , the wave number$\left(\frac{1}{\lambda }\right)$is given by

$\frac{1}{\lambda }= Z^{2}R \left(\frac{1}{n_1^2}-\frac{1}{n_2^{ 2}}\right)$

where Z is atomic number, R the Rydberg's constant and n is the quantum number.

$\Rightarrow \lambda \propto \frac{1}{{\text{Z}}^2}$

Atomic number of lithium is 3, of helium is 2 and of hydrogen is 1.

$\therefore {\lambda }_{{\text{Li}}^{2+}} : {\lambda }_{{\text{He}}^{+}} : {\lambda }_{\text{H}}=\frac{1}{{\left(3\right)}^2} : \frac{1}{{\left(2\right)}^2}\text{ : }1$

$=\frac{1}{9}\text{ : }\frac{1}{4}\text{ : }1$