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An electron has a mass of 9.1 10⁻³¹ ~kg . It revolves round the nucleus in a circular orbit of radius 0.529 10⁻¹⁰ ~m at a speed of 2.2 10^6 ~ms ⁻¹ . The magnitude of its angular momentum is

Options

  1. A2.06 10⁻³⁴ Kg m ^2 ~s ⁻¹
  2. B1.06 10⁻³⁴ Kg m ^2 ~s ⁻¹
  3. C2.06 10⁻²⁴ Kg m ^2 ~s ⁻¹
  4. D1.06 10⁻²⁴ Kg m ^2 ~s ⁻¹

Correct answer

B. 1.06 10⁻³⁴ Kg m ^2 ~s ⁻¹

Step-by-step solution

The angular momentum L of a particle moving in a circular orbit is given by the formula L = mvr , where m is the mass of the particle, v is its speed, and r is the radius of the orbit. Given values are: m = 9.1 10⁻³¹ kg v = 2.2 10⁶ ms ⁻¹ r = 0.529 10⁻¹⁰ m Substituting these values into the formula: L = (9.1 10⁻³¹) (2.2 10⁶) (0.529 10⁻¹⁰) Calculating the product of the numerical coefficients: 9.1 2.2 0.529 = 20.02 0.529 10.59 10.6 Calculating the power of 10: 10⁻³¹ 10⁶ 10⁻¹⁰ = 10^ -31 + 6 - 10 = 10⁻³⁵ Combining thes

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