Question

Angular momentum of electron in the second orbit of hydrogen atom is :

• A. \( 1.05 \times 10^{-34} \, \text{J} \cdot \text{s} \)
• B. \( 1.05 \times 10^{-36} \, \text{J} \cdot \text{s} \)
• C. \( 2.1 \times 10^{-34} \, \text{J} \cdot \text{s} \)
• D. \( 2.1 \times 10^{-31} \, \text{J} \cdot \text{s} \)

Hint:

The angular momentum of an electron in a Bohr orbit is quantized and given by \( L = n \hbar \), where \( n \) is the principal quantum number.

Answer 1

The Correct Option is C

Step 1: Angular Momentum of Electron.
The angular momentum \( L \) of an electron in a Bohr orbit is quantized and given by the equation: \[ L = n \hbar \] where \( n \) is the principal quantum number and \( \hbar \) is the reduced Planck's constant (\( \hbar = \frac{h}{2\pi} \)). For the second orbit (\( n = 2 \)) in the hydrogen atom, the angular momentum is: \[ L = 2 \times \frac{h}{2\pi} \] Substituting \( h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \), we get: \[ L = 2 \times \frac{6.626 \times 10^{-34}}{2\pi} = 2.1 \times 10^{-34} \, \text{J} \cdot \text{s} \]
Step 2: Conclusion.
Thus, the angular momentum of the electron in the second orbit is \( 2.1 \times 10^{-34} \, \text{J} \cdot \text{s} \), corresponding to option .