Question

electronic tranisition that could lead to absorbtion of light by atom is
either 1S-4P-3P-5F-4S-2P ??

Answer 1

The energy states of electrons are labeled by a number n which is called principal quantum number. The number of the ground state is n=1, that of the first excited state is n=2, and so on. The letters s, p, d, f written after the principal quantum number stand for the azimuthal quantum number which is the quantum number of orbital angular momentum, and s means l = 0, p means l = 1, d stands for l = 2, f for l = 3, and then it continues with the alphabet.

For an electron at the ground state (n = 1) to be moved up to the next level (n = 2) it must absorb a quantum of energy that is the perfect amount to make this move. If the quantum is too small the electron could not reach the next level, so it doesn't try. If the quantum is too large the electrons would overshoot the next level, so again, it does not try. Only quanta of exactly the right size will be absorbed and used.

Similarly, if an electron is already at the second level (n = 2), and there is a space for the electron at the lower level (n = 1), it can release a quantum of energy and drop down to the lower level. But the amount of energy given off will be a whole number quantum. If this energy is given off as light then the photons rushing away from the falling electron will be of only one size and color.

All this implies that if white light is shone on certain materials or substances only certain wavelengths will be absorbed by the electrons in that substance.

Therefore, the transition of an electron from one energy state to another can be labeled as 1s-4p or 3p-5f and so on.