Question

The energy of an electron in an orbit of Bohr hydrogen atom is \( -3.4 \, \text{eV} \). Find its angular momentum.

Hint:

In Bohr's model, the angular momentum of an electron in the nth orbit is quantized and given by \( L = n \hbar \), where \( n \) is the principal quantum number.

Answer 1

1. Energy of the Electron in Bohr Hydrogen Atom:

The energy of an electron in the \( n \)-th orbit of a Bohr hydrogen atom is given by the equation:

\[ E_n = - \frac{13.6}{n^2} \, \text{eV} \]

Where:

• \( E_n \) is the energy of the electron in the \( n \)-th orbit (in eV).
• 13.6 eV is the energy of the electron in the first orbit (n = 1).
• \( n \) is the principal quantum number (for the first orbit, \( n = 1 \)).

 

For \( n = 1 \), the energy of the electron is \( E_1 = -13.6 \, \text{eV} \), which is the energy in the ground state. The given energy is \( -3.4 \, \text{eV} \), so we can infer that the electron is in the second orbit where \( n = 2 \). This is confirmed by the following calculation:

\[ E_2 = - \frac{13.6}{2^2} = - \frac{13.6}{4} = - 3.4 \, \text{eV} \]

2. Bohr’s Quantization Condition for Angular Momentum:

According to Bohr’s model, the angular momentum \( L \) of the electron in the \( n \)-th orbit is quantized and is given by:

\[ L = n \hbar \]

Where:

• \( L \) is the angular momentum of the electron.
• \( n \) is the principal quantum number (in this case, \( n = 2 \) for the second orbit).
• \( \hbar \) is the reduced Planck's constant, \( \hbar = \frac{h}{2\pi} \), where \(h = 6.626 \times 10^{-34} \, \text{J.s}\)

 

3. Calculating the Angular Momentum:

Substitute \(n = 2\)and \(\hbar = \frac{6.626 \times 10^{-34}}{2\pi} \, \text{J.s}\) into the equation for angular momentum:

\(L = 2 \times \frac{6.626 \times 10^{-34}}{2\pi}\) \(= 2.11 \times 10^{-34} \, \text{J.s}\)

4. Conclusion:

• The angular momentum of the electron in the second orbit of the Bohr hydrogen atom is \(L=2.11 \times 10^{-34} \, \text{J.s}\).