Question

An electron has a mass of $ 9.1 \times 10^{-31} \, \text{kg} $. It revolves round the nucleus in a circular orbit of radius $ 0.529 \times 10^{-10} \, \text{m} $ at a speed of $ 2.2 \times 10^6 \, \text{m/s} $. The magnitude of its angular momentum is

• A. \( 1.06 \times 10^{-34} \, \text{Kg m}^2 \text{s}^{-1} \)
• B. \( 1.06 \times 10^{-24} \, \text{Kg m}^2 \text{s}^{-1} \)
• C. \( 2.06 \times 10^{-34} \, \text{Kg m}^2 \text{s}^{-1} \)
• D. \( 2.06 \times 10^{-24} \, \text{Kg m}^2 \text{s}^{-1} \)

Hint:

The angular momentum for an electron in a circular orbit is the product of its mass, velocity, and radius. Ensure to substitute the correct values for each variable.

Answer 1

The Correct Option is A

The angular momentum \( L \) of an electron in a circular orbit is given by the formula: \[ L = m \cdot v \cdot r \] Where: - \( m = 9.1 \times 10^{-31} \, \text{kg} \) is the mass of the electron, - \( v = 2.2 \times 10^6 \, \text{m/s} \) is the speed of the electron, - \( r = 0.529 \times 10^{-10} \, \text{m} \) is the radius of the orbit. Substitute the given values: \[ L = (9.1 \times 10^{-31}) \cdot (2.2 \times 10^6) \cdot (0.529 \times 10^{-10}) \] \[ L = 1.06 \times 10^{-34} \, \text{Kg m}^2 \text{s}^{-1} \]
Thus, the magnitude of the angular momentum is \( 1.06 \times 10^{-34} \, \text{Kg m}^2 \text{s}^{-1} \).