In a H-like sample, electrons make transition from 4th excited state up to 2nd state. Then

It seems like you didn't finish your question. However, I can start explaining based on what you've given so far.

In a hydrogen-like atom (H-like), electrons can transition between energy levels. When an electron moves from a higher energy level to a lower one, it releases energy in the form of light (a photon).

The energy of the photon (and hence the color of the light) depends on the difference in energy between the two levels.

In your case, the electron is moving from the 4th excited state to the 2nd state. The 4th excited state is actually the 5th energy level (n=5), because we start counting from the ground state (n=1). The 2nd state is the 3rd energy level (n=3).

The energy difference between these two levels can be calculated using the formula for the energy levels of a hydrogen-like atom:

E = -13.6 eV * (1/n^2)

where E is the energy, n is the energy level, and eV (electron volts) is the unit of energy.

The energy of the 5th level is -13.6 eV * (1/5^2) = -0.544 eV, and the energy of the 3rd level is -13.6 eV * (1/3^2) = -1.51 eV.

The energy of the photon released is the difference between these two energies, which is -1.51 eV - (-0.544 eV) = -0.966 eV.

This energy corresponds to a photon in the infrared region of the spectrum.

Please complete your question if you need more information.