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 =6.022xx10^23)

The ionization energy of hydrogen atom is -13.6 eV. The energy require

1.	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 =6.022xx10^23)
Answer» `1.69xx10^(-20)` J `1.69xx10^(-23)` J `1.69xx10^(23)` J `1.69xx10^(25)` J Solution :`E=(-13.6)/n^2 =(-13.6)/4 =-3.4 EV` We know that energy required for excitation `DeltaE=E_2-E_1=-3.4-(-13.6)=10.2 eV` Therefore energy required for excitation of ELECTRON per atom `=10.2/(6.02xx10^23)=1.69xx10^(-23)` J