Question

In Bohr model of hydrogen atom, an electron is revolving in second orbit. Find the value of:
(i) angular momentum of electron,
(ii) radius of the orbit, and
(iii) kinetic energy of electron.
Take radius of first orbit of hydrogen atom as 0.5 \AA.

Hint:

In the Bohr model, the electron's energy, angular momentum, and radius depend on the quantum number \( n \). For each orbit, the angular momentum is quantized, and the electron's energy is negative, indicating that the electron is bound to the nucleus.

Answer 1

In the Bohr model of the hydrogen atom, the electron in the \( n \)-th orbit has the following properties: 1. Angular Momentum: According to the Bohr model, the angular momentum \( L \) of the electron in the \( n \)-th orbit is quantized and given by: \[ L = n \hbar \] where \( n \) is the principal quantum number and \( \hbar = \frac{h}{2 \pi} \) is the reduced Planck's constant. For \( n = 2 \) (second orbit), we have: \[ L = 2 \hbar \] Using the value of \( \hbar = 1.055 \times 10^{-34} \, \text{J·s} \), we get: \[ L = 2 \times 1.055 \times 10^{-34} \, \text{J·s} = 2.11 \times 10^{-34} \, \text{J·s} \] 2. Radius of the Orbit: The radius \( r_n \) of the \( n \)-th orbit is given by the formula: \[ r_n = n^2 r_1 \] where \( r_1 \) is the radius of the first orbit. Given \( r_1 = 0.5 \, \text{\AA} = 0.5 \times 10^{-10} \, \text{m} \) and \( n = 2 \), the radius of the second orbit is: \[ r_2 = 2^2 \times 0.5 \times 10^{-10} \, \text{m} = 4 \times 0.5 \times 10^{-10} \, \text{m} = 2 \times 10^{-10} \, \text{m} \] 3. Kinetic Energy of the Electron: The kinetic energy \( K \) of the electron in the \( n \)-th orbit is related to the total energy \( E \) by: \[ K = \frac{1}{2} m v^2 \] where \( m \) is the mass of the electron and \( v \) is the speed of the electron. The total energy \( E \) of the electron in the \( n \)-th orbit is: \[ E = -\frac{k e^2}{2r_n} \] where \( k = 9 \times 10^9 \, \text{N·m}^2/\text{C}^2 \) and \( e = 1.6 \times 10^{-19} \, \text{C} \) is the charge of the electron. The kinetic energy is the negative of half of the total energy: \[ K = -\frac{E}{2} = \frac{k e^2}{4 r_n} \] Substituting \( r_n = 2 \times 10^{-10} \, \text{m} \) into the formula, we get: \[ K = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{4 \times 2 \times 10^{-10}} \, \text{J} \] After calculating, we find: \[ K \approx 1.15 \times 10^{-18} \, \text{J} \] Thus, the answers are: \begin{enumerate} \item Angular momentum: \( L = 2.11 \times 10^{-34} \, \text{J·s} \) \item Radius of the orbit: \( r_2 = 2 \times 10^{-10} \, \text{m} \) \item Kinetic energy of the electron: \( K \approx 1.15 \times 10^{-18} \, \text{J} \) \end{enumerate}