A hydrogen atom in the ground state is excited by an electron beam of 12.5 eV energy . Find out the maximum number of lines emitted by the atom from its excited state.

Explanation

We know that energy of an electron in nth orbit of hydrogen is `E_(n) = - (13.6)/(n^(2)) e V` and energy is ground state of hydrogen is - 13.6 eV.
When an electron beam of 12.5 eV is used , hydrogen atom cannot be ionised but at the most excited to n = 3 state where energy of the electron will be ` - (13.6)/((3)^(2)) = - 1.51 eV` (as maximum energy of electron can be ` - 13.6 + 12.5 = - 1.1 eV` and electron cannot be excited to n = 4 state having energy - 0.85 e V).
So, only three transitions are possible as show here . Here transitions No.1 and 2 represent lines of Lyman series and transition No.2 represents line of Balmer series .