A sample of hydrogen gas has same atom in out excited state and same atom in other excited state it emits three difference photon.When the sample was irradiated with radiation of energy `2.85 eV` ,it emits `10` different photon all having energy in or less than `13.6 eV` ltbrtgt a. Find the principal quantum number of initially excited electrons
b. Find the maximum and minimum energies of the initially emitted photon

Explanation

Initailly the sample emit three difference photon it means for one of the ecxited states `n = 3`
For the other excited state `n = 1 or n = 2`
After transition it emit `10` diofference photon it means for the excited state `n = 5`
Energy of fifth energy level `= (-13.6)/((5)^(2)) = - 0.544 eV`
Energy of total energy level `= (-13.6)/((3)^(2)) = - 1.51 eV`
Energy of second energy level `= (-13.6)/((2)^(2)) = - 3.4 eV`
So, after absurption of `2.85 eV` energy only electron from accond energy level can jump to fifth energy level principal quantum number of initially excited state are `2` and `3` respectively
Maximum energy of emitted photon `= 13.6 - 1.51 = 12.09 eV`
Maximum energy of emitted photon `= 3.4 - 1.51 = 1.89 eV`