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=K536For removing an electron n1 = 1 to n2=∞Energy  E1=K[1]=365E=7.2 E∴Ionization energy = 7.2 E