Step-by-step Solution:

1. Determine the principal quantum number (n):

The orbital angular momentum (L) of an electron in a hydrogen-like atom is quantized according to Bohr's model: $L = n \frac{h}{2\pi}$ where n is the principal quantum number and h is Planck's constant.

It is given that the orbital angular momentum is: $L = \frac{3h}{2\pi}$

Comparing the two expressions for L, we get: $n \frac{h}{2\pi} = \frac{3h}{2\pi}$ $n = 3$ So, the electron is in the third orbit (second excited state).

2. Determine the atomic number (Z):

The radius of the n-th orbit ($r_n$) in a hydrogen-like atom is given by: $r_n = a_0 \frac{n^2}{Z}$ where $a_0$ is the Bohr radius and Z is the atomic number.

It is given that the radius of the orbit is $r = 4.5 a_0 = \frac{9}{2} a_0$. Since we found n=3, this is the radius of the third orbit, $r_3$.

Substituting the values of n and $r_3$ into the radius formula: $\frac{9}{2} a_0 = a_0 \frac{3^2}{Z}$ $\frac{9}{2} = \frac{9}{Z}$ $Z = 2$ The hydrogen-like atom is a singly ionized Helium atom (He⁺).

3. Identify possible de-excitation transitions:

The electron is in the n=3 state. It can de-excite to lower energy levels. The possible transitions are:

• $n = 3 \rightarrow n = 1$ (direct transition to ground state)
• $n = 3 \rightarrow n = 2$ (first step of a cascade)
• $n = 2 \rightarrow n = 1$ (second step of a cascade, after the $3 \rightarrow 2$ transition)

We need to find the wavelengths of the photons emitted during these transitions.

4. Calculate the wavelengths using the Rydberg formula:

The Rydberg formula for a hydrogen-like atom is: $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$ where R is the Rydberg constant, $n_1$ is the final state, and $n_2$ is the initial state ($n_2 > n_1$). Here, we have Z=2.

• For the transition $n=3 \rightarrow n=1$: $\frac{1}{\lambda_1} = R (2^2) \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = 4R \left( 1 - \frac{1}{9} \right) = 4R \left( \frac{8}{9} \right) = \frac{32R}{9}$ $\lambda_1 = \frac{9}{32R}$

• For the transition $n=3 \rightarrow n=2$: $\frac{1}{\lambda_2} = R (2^2) \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 4R \left( \frac{1}{4} - \frac{1}{9} \right) = 4R \left( \frac{9-4}{36} \right) = 4R \left( \frac{5}{36} \right) = \frac{5R}{9}$ $\lambda_2 = \frac{9}{5R}$

• For the transition $n=2 \rightarrow n=1$: $\frac{1}{\lambda_3} = R (2^2) \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 4R \left( 1 - \frac{1}{4} \right) = 4R \left( \frac{3}{4} \right) = 3R$ $\lambda_3 = \frac{1}{3R}$

The possible wavelengths of emitted photons during de-excitation are 9/(32R), 9/(5R), and 1/(3R).

5. Compare with the given options:

• A: $\frac{9}{32R}$ - This matches our calculated wavelength $\lambda_1$. This is a correct option.
• B: $\frac{9}{16R}$ - This does not match any of our calculated wavelengths.
• C: $\frac{9}{5R}$ - This matches our calculated wavelength $\lambda_2$. This is a correct option.
• D: $\frac{4}{3R}$ - This does not match any of our calculated wavelengths.

Thus, the possible wavelengths are given in options A and C.