how many e V are absorbed by an electron from the L alpha photon l_alpha photon (la) when the electron is excited from its ground stated (n=1) to the first excited level (n=2)
To find out how many electron volts (eV) are absorbed by an electron when it is excited from its ground state (n=1) to the first excited level (n=2) by a specific photon, in this case, the L alpha photon (l_alpha), we need to understand the energy difference between these two energy levels.
The energy of an electron in a hydrogen-like atom can be calculated using the formula:
E = -13.6 eV / n^2
where E is the energy, -13.6 eV represents the ionization energy of hydrogen, and n represents the principal quantum number.
For the ground state (n=1), we have:
E1 = -13.6 eV / (1^2) = -13.6 eV
For the first excited level (n=2), we have:
E2 = -13.6 eV / (2^2) = -3.4 eV
The energy difference between these two levels is:
ΔE = E2 - E1 = (-3.4 eV) - (-13.6 eV) = 10.2 eV
Therefore, when the electron is excited from the ground state to the first excited level by the L alpha photon, it absorbs 10.2 electron volts (eV) of energy.