• # question_answer The ionization energy of hydrogen atom is $-13.6\,eV.$ The energy required to excite the electron in a hydrogen atom from the ground state to the first excited state is (Avogadro?s constant = 6.022 × 1023)   [BHU 1999] A)                 $1.69\times {{10}^{-20}}J$         B)                 $1.69\times {{10}^{-23}}J$ C)                 $1.69\times {{10}^{23}}J$          D)                 $1.69\times {{10}^{25}}J$

$E=\frac{-13.6}{{{n}^{2}}}=\frac{-13.6}{4}=-3.4\,eV$ We know that energy required for excitation $\Delta E={{E}_{2}}-{{E}_{1}}$ $=-3.4-(-13.6)=10.2\,eV$                 Therefore energy required for excitation of electron per atom $=\frac{10.2}{6.02\times {{10}^{23}}}=1.69\times {{10}^{-23}}J$