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`)

Options

A. `1.69xx10^(-20)` J
B. `1.69xx10^(-23)` J
C. `1.69xx10^(23)` J
D. `1.69xx10^(25)` J

Answer

B

Explanation

`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