Question

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

• A. 1.275 MeV
• B. 12.75 eV
• C. 12.75 J
• D. 12.75 MeV

Hint:

When calculating the energy difference between two energy levels of a hydrogen atom, use the formula \( E_n = - \frac{13.6}{n^2} \, \text{eV} \) and take the difference between the higher and lower energy states.

Answer 1

The Correct Option is B

The energy levels of a hydrogen atom are given by the formula: \[ E_n = - \frac{13.6}{n^2} \, \text{eV} \] where \( n \) is the principal quantum number of the energy level. - The ground state energy corresponds to \( n = 1 \), so: \[ E_1 = - \frac{13.6}{1^2} = -13.6 \, \text{eV} \] - The energy of the third excited state corresponds to \( n = 4 \), so: \[ E_4 = - \frac{13.6}{4^2} = - \frac{13.6}{16} = -0.85 \, \text{eV} \] Now, the energy of the radiation emitted is the difference between the energy of the third excited state and the ground state: \[ \Delta E = E_1 - E_4 = (-13.6) - (-0.85) = -13.6 + 0.85 = -12.75 \, \text{eV} \] Therefore, the energy of the radiation emitted is \( 12.75 \, \text{eV} \).