Question

The energy of a hydrogen atom in an energy state is \( -3.4 \) electron-volts. For an electron in this energy state, calculate the following: (i) Binding energy
(ii) Ionization potential
(iii) The number of energy states
(iv) The angular momentum
(v) Kinetic energy
(i) Binding Energy

Hint:

For hydrogen atoms, ionization energy equals the binding energy in a given state.

Answer 1

Binding energy is the energy required to remove the electron from the given state. \[ \text{Binding energy} = 3.4 \text{ eV} \] \[ \boxed{3.4 \text{ eV}} \] (ii) Ionization Potential % Solution Solution: Ionization potential is the energy required to ionize the atom by removing an electron from the current state. \[ \text{Ionization potential} = 3.4 \text{ eV} \] \[ \boxed{3.4 \text{ eV}} \] (iii) The Number of Energy States % Solution Solution: The number of possible states for an electron in the nth energy level is given by: \[ \text{Number of states} = n^2 \] For \( n = 2 \): \[ 2^2 = 4 \] \[ \boxed{4} \] (iv) The Angular Momentum % Solution Solution: The angular momentum of an electron in the nth orbit is: \[ L = n \hbar \] For \( n = 2 \): \[ L = 2\hbar \] \[ \boxed{2\hbar} \] (v) Kinetic Energy % Solution Solution: In hydrogen atoms, kinetic energy in an energy state is numerically equal to the binding energy. \[ \text{Kinetic Energy} = 3.4 \text{ eV} \] \[ \boxed{3.4 \text{ eV}} \]