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 described by the equation: \[ E_n = -\frac{13.6}{n^2} \, \text{eV} \] where \( n \) represents the principal quantum number. For the ground state, \( n = 1 \), yielding an electron energy of: \[ E_1 = -\frac{13.6}{1^2} = -13.6 \, \text{eV} \] For the first excited state, with \( n = 2 \), the electron's energy is: \[ E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4 \, \text{eV} \] The total energy \( E_2 \) is the sum of kinetic energy \( K \) and potential energy \( U \). According to the Virial Theorem: \[ K = -\frac{1}{2} U \] The total energy is given by: \[ E_2 = K + U \] Substituting \( K = -\frac{1}{2} U \) into the total energy equation results in \( E_2 = -\frac{1}{2} U + U = \frac{1}{2} U \). This implies \( U = 2 E_2 \). Given that \( E_2 = -3.4 \, \text{eV} \) for the first excited state, the potential energy is: \[ U = 2 \times (-3.4) = -6.8 \, \text{eV} \] Using the Virial Theorem relation \( K = -\frac{1}{2} U \), the kinetic energy is: \[ K = -\frac{1}{2} \times (-6.8) = 3.4 \, \text{eV} \] Therefore, 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