Question

The energy of an electron of excited hydrogen atom is -3.4 eV. Determine the angular momentum of the electron.

Hint:

The angular momentum of an electron in Bohr’s model is quantized and directly proportional to the principal quantum number \( n \).

Answer 1

According to Bohr's model, the energy of the electron in the \( n \)-th orbit is given by:
\[ E_n = - \frac{13.6 \, \text{eV}}{n^2} \] We are given the energy \( E = -3.4 \, \text{eV} \). To find \( n \), we solve for \( n \) in the equation:
\[ -3.4 = - \frac{13.6}{n^2} \] \[ n^2 = \frac{13.6}{3.4} = 4 \] \[ n = 2 \] Thus, the electron is in the second energy level (\( n = 2 \)).
The angular momentum \( L \) of the electron in the \( n \)-th orbit is given by:
\[ L = n \hbar \] where \( \hbar = \frac{h}{2\pi} \) is the reduced Planck’s constant.
For \( n = 2 \):
\[ L = 2 \hbar \] Thus, the angular momentum of the electron is:
\[ \boxed{L = 2 \hbar} \]