The ionization enthalpy of hydrogen atom is $1.312\times {10}^{6}Jmo{l}^{-1}$.The energy required to excite the electron in the atom from n = 1 to n = 2 is 

The energy difference between the two orbits of hydrogen atom is given by

                                                   $\Delta E=N_A{R}_{\infty }hc(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

            For ionization $n_1=1and{n}_2=\infty .Hence,N_A{R}_{\infty }hc=1.312\times {10}^{6}mo{l}^{-1}$

            For $n=1\to n=2,$the energy difference is

            $\Delta E=\frac{3}{4}\left(N_A{R}_{\infty }hc\right)=\frac{3}{4}\times 1.312\times {10}^{6}Jmo{l}^{-1}=9.84\times {10}^{5}Jmo{l}^{-1}$