The radius of the orbit of an electron in a hydrogen like atom is...

The orbital angular momentum of an electron is given by $L = \dfrac{nh}{2\pi}$.

Given $L = \dfrac{3h}{2\pi}$, we get $n = 3$.

The radius of the $n$-th orbit in a hydrogen-like atom is $r = \dfrac{n^2}{Z}a_0$.

Substituting $r = 4.5a_0 and n = 3$:

$4.5a_0 = \dfrac{3^2}{Z}a_0 \Rightarrow Z = \dfrac{9}{4.5} = 2$

The possible transitions from the $n = 3$ state to lower energy levels are $n = 3 \rightarrow 1 and n = 3 \rightarrow 2$.

Using the Rydberg formula for wavelength:

$\dfrac{1}{\lambda} = RZ^2 \left( \dfrac{1}{n_f^2} - \dfrac{1}{n_i^2} \right)<br><br>For the transition$ n = 3 → 1$:<br><br>\dfrac{1}{\lambda} =$ R(2)^2 \left( \dfrac{1}{1^2} - \dfrac{1}{3^2} \right) =4R \left(1 - \dfrac{1}{9} \right) =4R \left( \dfrac{8}{9} \right) = \dfrac{32R}{9}

$\lambda =$\dfrac{9}{32R}

For$the transition n = 3 → 2$:

\dfrac{1}{\lambda} =$R(2)^2 \left( \dfrac{1}{2^2} - \dfrac{1}{3^2} \right) =4R \left( \dfrac{1}{4} - \dfrac{1}{9} \right) =4R \left( \dfrac{5}{36} \right) = \dfrac{5R}{9}<br><br>$\lambda = $\dfrac{9}{5R}<br><br>Among$ the given options,$\dfrac{9}{32R}$ is a possible wavelength.

Answer: $\dfrac{9}{32R}$