Question:medium

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.

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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.
Updated On: Jan 13, 2026
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Solution and Explanation

The Bohr model of the hydrogen atom describes the electron in the \( n \)-th orbit with the following characteristics: 1. Angular Momentum: Quantized according to \( L = n \hbar \), where \( n \) is the principal quantum number and \( \hbar \) is the reduced Planck's constant (\( \hbar = \frac{h}{2 \pi} \)). For the second orbit (\( n = 2 \)), the angular momentum is \( L = 2 \hbar \). Using \( \hbar = 1.055 \times 10^{-34} \, \text{J·s} \), the calculated angular momentum is \( L = 2.11 \times 10^{-34} \, \text{J·s} \). 2. Orbit Radius: The radius \( r_n \) of the \( n \)-th orbit is given by \( r_n = n^2 r_1 \), where \( r_1 \) is the radius of the first orbit. With \( r_1 = 0.5 \, \text{\AA} = 0.5 \times 10^{-10} \, \text{m} \) and \( n = 2 \), the radius of the second orbit is \( r_2 = 2^2 \times 0.5 \times 10^{-10} \, \text{m} = 2 \times 10^{-10} \, \text{m} \). 3. Electron Kinetic Energy: The kinetic energy \( K \) is \( K = \frac{1}{2} m v^2 \), where \( m \) is the electron's mass and \( v \) is its speed. The total energy \( E \) in the \( n \)-th orbit is \( E = -\frac{k e^2}{2r_n} \), with \( k = 9 \times 10^9 \, \text{N·m}^2/\text{C}^2 \) and \( e = 1.6 \times 10^{-19} \, \text{C} \). Kinetic energy is \( K = -\frac{E}{2} = \frac{k e^2}{4 r_n} \). Substituting \( r_n = 2 \times 10^{-10} \, \text{m} \) yields \( K = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{4 \times 2 \times 10^{-10}} \, \text{J} \), which is approximately \( K \approx 1.15 \times 10^{-18} \, \text{J} \). Summary of values for the second orbit (\( n = 2 \)):
  1. Angular momentum: \( L = 2.11 \times 10^{-34} \, \text{J·s} \)
  2. Radius of the orbit: \( r_2 = 2 \times 10^{-10} \, \text{m} \)
  3. Kinetic energy of the electron: \( K \approx 1.15 \times 10^{-18} \, \text{J} \)
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