In hydrogen atom, if the difference in the energy of the electron in n=2 and n=3 orbits is E, the ionization energy of hydrogen atom is

Energy, E=K1n12−1n22(K= constant ) n1=2 and n2=3 . So , E=K122−132=K536  For removing an electron n1=1 to n2=∞  Energy E1=K[1]=365E=7.2E ∴ Ionization energy =7.2E