Question

An electron in an atom is revolving round the nucleus in a circular orbit of radius 5.3 × 10\(^{-11}\) m with a speed of 3 × 10\(^6\) m/s. Find the angular momentum of electron.

Hint:

Angular momentum is \( L = m v r \); ensure consistent SI units for mass, velocity, and radius.

Answer 1

Angular momentum \( L = m v r \), where \( m \) is the electron mass (\( 9.11 \times 10^{-31} \, \text{kg} \)), \( v = 3 \times 10^6 \, \text{m/s} \), \( r = 5.3 \times 10^{-11} \, \text{m} \).
\[ L = (9.11 \times 10^{-31}) \times (3 \times 10^6) \times (5.3 \times 10^{-11}). \] \[ = 9.11 \times 3 \times 5.3 \times 10^{-31 + 6 - 11} = 144.729 \times 10^{-36} = 1.44729 \times 10^{-34} \, \text{kg} \cdot \text{m}^2/\text{s}. \] Answer: \( 1.45 \times 10^{-34} \, \text{kg} \cdot \text{m}^2/\text{s} \) (to 3 significant figures).