In a hydrogen like atom, electron is in 2^(nd) excited state, the binding energy of fourth state of this atom is 13.6 eV, then

In a hydrogen like atom, electron is in 2^(nd) excited state, the bind

1.	In a hydrogen like atom, electron is in 2^(nd) excited state, the binding energy of fourth state of this atom is 13.6 eV, then
Answer» A 25 eV photon can set FREE the electron from the second excited state of this SAMPLE. 3 different types of photon will be observed if ELECTRONS make transition up to ground state from the second excited state If 23 eV photon is used then K.E. of the ejected electron is 1eV. `2^(nd)` line of balmer series of this sample has same energy VALUE as `1^(st)` excitation energy of H-atoms Solution :B.E. of `4^(th)` state `=13.6 (z^2)/(n^2) implies13.6 (z^2)/(4^2) = 13.6 impliesz= 4` Sample is `Be^(3+) therefore` energy of electron in `3^(rd)` state Therefore 25 eV photon will cause ionization