Question

The radius of an orbit in a Hydrogen - like atom is $4.5a_0$, where $a_0$ is the Bohr radius. Its orbital angular momentum is $\frac{3h}{2\pi }$, it is given that $h$ is Planck's constant and $R$ is Rydberg constant. The possible wavelength when the atom de-excites from the given state to the ground state is:

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

Answer 1

The correct option is A $\frac{9}{32R}$

Angular momentum $=\frac{nh}{2\pi }=\frac{3h}{2\pi }$
$\Rightarrow n=3$
Also $r=\frac{n^2}{Z}{a}_0$
$\Rightarrow \frac{n^2}{Z}{a}_0=\frac{9a_0}{2}=\frac{3^2{a}_0}{2}$
$\Rightarrow Z=2$
Now de-excitation is occuring from $n=3$
to Ground state; i.e $n=1$
For de-excitation, $\frac{1}{\lambda }=R{Z}^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
$\frac{1}{\lambda }=R\times Z^2\times (\frac{1}{1^2}-\frac{1}{3^2})$
$\frac{1}{\lambda }=4R\left(\frac{9-1}{9}\right)$
$\frac{1}{\lambda }=4R\times \frac{8}{9}$
$\frac{1}{\lambda }=\frac{32R}{9}$
$\lambda =\frac{9}{32R}$