Question

The energy of an electron in the ground state of hydrogen atom is -13.6 eV. The kinetic and potential energy of the electron in the first excited state will be:

• A. \( -13.6 \, \text{eV}, 27.2 \, \text{eV} \)
• B. \( -6.8 \, \text{eV}, 13.6 \, \text{eV} \)
• C. \( 3.4 \, \text{eV}, -6.8 \, \text{eV} \)
• D. \( 6.8 \, \text{eV}, -3.4 \, \text{eV} \)

Hint:

In the Bohr model of the hydrogen atom, the total energy is the sum of the kinetic and potential energies. The kinetic energy is equal in magnitude but opposite in sign to the potential energy.

Answer 1

The Correct Option is C

The energy of an electron in the hydrogen atom is given by the formula: \[ E_n = -\frac{13.6}{n^2} \, \text{eV} \] Where \( n \) is the principal quantum number. In the ground state, \( n = 1 \), so the energy of the electron in the ground state is: \[ E_1 = -\frac{13.6}{1^2} = -13.6 \, \text{eV} \] For the first excited state, \( n = 2 \), so the energy of the electron in the first excited state is: \[ E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4 \, \text{eV} \] Now, the total energy \( E_2 \) consists of both kinetic energy \( K \) and potential energy \( U \). According to the Virial Theorem: \[ K = -\frac{1}{2} U \] The total energy is: \[ E_2 = K + U \] Since \( K = -\frac{1}{2} U \), it follows that \( E_2 = -\frac{1}{2} U + U = \frac{1}{2} U \), so \( U = 2 E_2 \). For the first excited state, we already know that \( E_2 = -3.4 \, \text{eV} \). Therefore: \[ U = 2 \times (-3.4) = -6.8 \, \text{eV} \] And since \( K = -\frac{1}{2} U \): \[ K = -\frac{1}{2} \times (-6.8) = 3.4 \, \text{eV} \] Thus, the kinetic energy is \( 3.4 \, \text{eV} \) and the potential energy is \( -6.8 \, \text{eV} \). % Correct Answer Correct Answer:} (C) 3.4 eV, -6.8 eV