Question

The ground state energy of hydrogen atom is -13.6 eV. When its electron is in the first excited state, its excitation energy is

• A. 10.2eV
• B. 0
• C. 3.4eV
• D. 6.8eV

Answer 1

The Correct Option is A

The problem involves finding the excitation energy when the electron in a hydrogen atom transitions from its ground state to the first excited state. Let's solve this step-by-step.

1. To solve this, we must first understand the energy levels of a hydrogen atom. The energy levels \(E_n\) 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.
2. The ground state of hydrogen corresponds to \(n = 1\). So, E_1 = -13.6 \, \text{eV}.
3. The first excited state corresponds to \(n = 2\). So, E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4 \, \text{eV}.
4. The excitation energy is the difference between the energy of the higher (excited) state and the lower (ground) state: \Delta E = E_2 - E_1.
5. Substituting the known values: \Delta E = (-3.4) - (-13.6) = 13.6 - 3.4 = 10.2 \, \text{eV}.

Thus, the excitation energy when the electron in a hydrogen atom moves from the ground state to the first excited state is 10.2 eV.