Given information:
- Laser line used: 20491 cm-1
- First Stokes line: 20479 cm-1
- Oxygen molecule contains only 16O

Explanation:
The rotational Raman spectrum occurs when a molecule undergoes a rotational transition due to the interaction with photons. In this case, the oxygen molecule is excited by the laser line at 20491 cm-1.

Rotational Raman effect:
The rotational Raman effect occurs when a linearly polarized incident photon interacts with a molecule and undergoes a change in its energy and polarization. The energy change is determined by the rotational energy levels of the molecule.

Raman scattering:
In Raman scattering, the scattered light can have a higher energy (anti-Stokes) or lower energy (Stokes) than the incident light. The energy difference corresponds to the change in rotational energy of the molecule.

First Stokes line:
The first Stokes line in the rotational Raman spectrum corresponds to a lower energy (longer wavelength) than the incident laser line. In this case, the first Stokes line is observed at 20479 cm-1, which is lower than the incident laser line at 20491 cm-1.

Explanation of expected next Stokes line:
The next rotational Stokes line can be determined by considering the energy difference between the incident laser line and the first Stokes line. The energy difference between the laser line (20491 cm-1) and the first Stokes line (20479 cm-1) is 12 cm-1.

The rotational energy levels of a diatomic molecule can be approximated by the expression:
E(J) = BeJ(J+1) - αeJ(J+1)^2

Where:
- E(J) is the rotational energy level
- B is the rotational constant
- J is the rotational quantum number
- α is the centrifugal distortion constant

Prediction of the next Stokes line:
To predict the next Stokes line, we need to find the change in rotational energy that corresponds to a difference of 12 cm-1. We can do this by comparing the energy differences for different rotational quantum numbers (J and J+1).

By comparing the energy differences, it can be observed that the change in energy for a difference of 1 in J is given by:
ΔE = 2B(J+1) - 2α(J+1)^2

Substituting J = 1, we get:
ΔE = 2B(2) - 2α(2)^2
ΔE = 4B - 8α

Since we know the energy difference (ΔE) is 12 cm-1, we can equate it to 12 cm-1 and solve for B and α.

12 = 4B - 8α

Calculation:
Given that the oxygen molecule contains only 16O, we can use the known values for B and α for oxygen.

For oxygen (16O), the rotational constants are:
B = 1.438 cm-1
α = 0.010 cm-1

Substituting these values into the equation, we get:
12 = 4(1.438) - 8(0.010)

Simplifying the equation gives:
12 = 5