The ground state energy of hydrogen atom is -13.6 ~eV . If the electron jumps from the 3^ rd excited state to the ground state then the energy of the radiation emitted will be:

Step-by-step solution

The energy of an electron in the n^ th orbit of a hydrogen atom is given by E_n = - 13.6 n^2 eV . The ground state corresponds to n = 1 . The 3^ rd excited state corresponds to n = 1 + 3 = 4 . The energy of the electron in the ground state is E₁ = - 13.6 1^2 = -13.6 eV . The energy of the electron in the 3^ rd excited state is E₄ = - 13.6 4^2 = - 13.6 16 = -0.85 eV . When the electron jumps from the 3^ rd excited state ( n=4 ) to the ground state ( n=1 ), the energy of the emitted radiation is given by E = E₄ - E₁