Q.In Raman effect, the wavelength of the incident radiation is 5890 Å. The wavelengths of Stokes' and anti-Stokes' lines are respectively :

Stokes lines are red-shifted (longer wavelength) and anti-Stokes lines are blue-shifted (shorter wavelength) relative to the incident radiation, so for 5890 Å incident light the Stokes/anti-Stokes pair is 5900 Å / 5880 Å.

In the Raman effect, monochromatic light incident on a molecular sample is scattered, and most of it (the Rayleigh line) leaves with unchanged wavelength (elastic scattering). A small fraction, however, is scattered inelastically as it exchanges energy with the vibrational/rotational energy levels of the molecule, producing symmetric pairs of weaker lines on either side of the Rayleigh line, called Raman lines.

When a photon gives up a quantum of energy $\Delta E$ to the molecule (exciting it, typically from its vibrational ground state to a higher vibrational level), the scattered photon leaves with less energy than it arrived with. Since photon energy $E=hc/\lambda$ is inversely proportional to wavelength, a lower-energy scattered photon corresponds to a longer wavelength than the incident light. These lower-frequency, longer-wavelength lines are called Stokes lines.

Conversely, if the molecule was already in an excited vibrational state before the collision, it can transfer some of its energy to the photon during scattering, leaving the scattered photon with more energy, and hence a shorter wavelength, than the incident light. These are called anti-Stokes lines. Because thermally excited molecules are always fewer in number than ground-state molecules (Boltzmann distribution), anti-Stokes lines are characteristically weaker in intensity than Stokes lines, though this intensity difference isn't what's being asked here — only the wavelength shift.

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