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. Electron Energy in Bohr Hydrogen Atom:

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

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

Where:

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

For \( n = 1 \), \( E_1 = -13.6 \, \text{eV} \), which is the ground state energy. Given an energy of \( -3.4 \, \text{eV} \), the electron is in the second orbit (\( n = 2 \)). This is validated by:

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

2. Bohr’s Angular Momentum Quantization:

Bohr’s model states that the angular momentum \( L \) of an electron in the \( n \)-th orbit is quantized as:

\[ L = n \hbar \]

Where:

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

 

3. Angular Momentum Calculation:

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

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

4. Result:

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