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

The energy of an electron in the nth orbit of the Bohr hydrogen atom is given by: \[ E_n = - \frac{13.6 \, \text{eV}}{n^2} \] For the first orbit, \( n = 1 \), and the energy is: \[ E_1 = -13.6 \, \text{eV} \] We are given that the energy of the electron is \( -3.4 \, \text{eV} \). This corresponds to the second orbit, where \( n = 2 \). The angular momentum \( L \) of the electron in the nth orbit is given by: \[ L = n \hbar \] where \( \hbar = \frac{h}{2\pi} \) is the reduced Planck's constant and \( n \) is the principal quantum number. For \( n = 2 \), the angular momentum is: \[ L = 2 \hbar \] Thus, the angular momentum of the electron is \( 2 \hbar \).