If radiation correcsponding to second line of “Balmer series” of Li^(2+) ion, knocked out electron from first excited state of H-atom, then kinetic energy of ejected electron would be:

If radiation correcsponding to second line of “Balmer series” of L

1.	If radiation correcsponding to second line of “Balmer series” of Li^(2+) ion, knocked out electron from first excited state of H-atom, then kinetic energy of ejected electron would be:
Answer» 2.55 eV 4.25 eV 11.25 eV 19.55 eV Solution :`E = 13.6 XX Z^2[1/(n_1^2)- 1/(n^2)] = 13.6xx3^2 [1/(2^2) - 1/(4^2)]`