Question

The ground state energy of a hydrogen atom is \( -13.6 \) eV. The potential energy of the electron in the first excited state of hydrogen is?

• A. \( -6.8 \) eV
• B. \( -3.4 \) eV
• C. \( -13.6 \) eV
• D. \( -27.2 \) eV

Hint:

In the hydrogen atom, the potential energy is always twice the total energy: \( U = 2E \). For the first excited state (\( n = 2 \)), the total energy is \( -3.4 \) eV, leading to a potential energy of \( -6.8 \) eV.

Answer 1

The Correct Option is A

Step 1: Understanding energy levels of the hydrogen atom 
The total energy of an electron in the \( n \)th energy level of a hydrogen atom is given by: \[ E_n = \frac{E_1}{n^2} \] where: - \( E_1 = -13.6 \) eV (ground state energy), - \( n \) is the principal quantum number. For the first excited state, \( n = 2 \): \[ E_2 = \frac{-13.6}{2^2} = \frac{-13.6}{4} = -3.4 \text{ eV} \] Step 2: Relationship between total energy and potential energy 
The potential energy (\( U \)) in an atomic system is related to the total energy by: \[ U = 2E_n \] Substituting \( E_2 = -3.4 \) eV: \[ U = 2 \times (-3.4) \] \[ U = -6.8 \text{ eV} \] Thus, the potential energy of the electron in the first excited state is \( -6.8 \) eV.