The radius of the orbit of an electron in a Hydrogen-like atom is 4.5 a₀, where a₀ is the Bohr radius. Its orbital angular momentum is $\frac{3\text{h}}{2\text{π}}\mathrm{.}$ It is given that h is Planck's constant and R is Rydberg constant. the possible wavelength (s), when the atom de-excites, is (are)

• Correct answer
  A

  $\frac{9}{32\text{R}}$
• B

  $\frac{9}{16\text{R}}$
• Correct answer
  C

  $\frac{9}{5\text{R}}$
• D

  $\frac{4}{3\text{R}}$

Solution

${\text{R}}_{\text{n}}=\text{4.5}{\text{a}}_0 \Rightarrow {\text{a}}_0{\text{n}}^2/\text{z}=\text{4.5 }{\text{a}}_0$
$\therefore \text{n}=3, \text{z}=2$
$\text{L}=\text{mvr}=\frac{3\text{h}}{2\pi }\left[\text{as n}=3\text{,}\text{z}=2\right]$

$\frac{1}{\lambda }={\text{Rz}}^2\left(\frac{1}{{\text{n}}_{\text{f}}^2}-\frac{1}{{\text{n}}_{\text{1}}^2}\right)$
$\frac{1}{{\lambda }_{3\to 1}}=\text{R}4\left[\frac{1}{1}-\frac{1}{9}\right]=4\text{R}\frac{8}{9}\Rightarrow {\lambda }_{3\to 1}=\frac{9}{32\text{R}}$
$\frac{1}{{\lambda }_{2\to 1}}=\text{R}4\left[\frac{1}{1}-\frac{1}{4}\right]=\frac{3}{4}4\text{R}\Rightarrow {\lambda }_{2\to 1}=\frac{1}{3\text{R}}$
$\frac{1}{{\lambda }_{3\to 2}}=\text{R}4\left[\frac{1}{4}-\frac{1}{9}\right]=\frac{5}{36}4\text{R}\Rightarrow {\lambda }_{3\to 2}=\frac{9}{5\text{R}}$

Hence, the correct answer is options: 'A', 'C'.