Question

The radius of the orbit of an electron in a Hydrogen-like atom is $4.5\,a_0$ where $a_0$ is the Bohr radius. Its orbital angular momentum is $\frac{3h}{2\pi}$ It is given that h is Planck constant and $R$ is Rydberg constant. The possible wavelength(s), when the atom de-excites, is (are)

• A. $\frac{9}{32R}$
• B. $\frac{9}{16R}$
• C. $\frac{9}{5R}$
• D. $\frac{4}{3R}$

Answer 1

Answer: (A), (C)

$L=3\bigg(\frac{h}{2\pi}\bigg)$
$\therefore n=3, as L=n\bigg(\frac{h}{2\pi}\bigg)$
$r_n \propto \frac{n^2}{Z}$
$r_3=4.5a_0$
$\therefore $ Z=2
$\frac{1}{\lambda_1}=Rz^2\bigg(\frac{1}{2^2}-\frac{1}{3^2}\bigg)=4R\bigg(\frac{1}{4}-\frac{1}{9}\bigg)$
$\therefore \lambda_1=\frac{9}{5R}$
$\frac{1}{\lambda_1}=Rz^2\bigg(\frac{1}{1^2}-\frac{1}{3^2}\bigg)=4R\bigg(1-\frac{1}{9}\bigg)$
$\Rightarrow \lambda_2=\frac{9}{32R}$
$\frac{1}{\lambda_3}=\bigg(\frac{1}{1^2}-\frac{1}{2^2}\bigg)=4R\bigg(1-\frac{1}{4}\bigg)$
$\Rightarrow =\frac{1}{3R}$