Question

The angular momentum of the electron revolving in the 2^nd orbit is

• A. \(\frac{h}{\pi}\)
• B. \(\frac{h}{2\pi}\)
• C. \(\frac{2h}{\pi}\)
• D. \(\frac{3h}{2\pi}\)
• E. \(\frac{h}{3\pi}\)

Hint:

The angular momentum of an electron in the n^{th} orbit is directly proportional to the principal quantum number n . For the 2^{nd} orbit, n = 2 , which simplifies the calculation.

Answer 1

The Correct Option is A

The angular momentum of an electron in an atomic orbital is quantized and given by the formula:

L = n * (h / 2π)

where:

• L is the angular momentum,
• n is the principal quantum number (which corresponds to the energy level or shell),
• h is Planck's constant (h = 6.626 × 10^-34 J·s).

For the 2^nd orbit, the principal quantum number n = 2. Substituting n = 2 into the formula:

L = 2 * (h / 2π)

Simplify the expression:

L = (2h / 2π) = (h / π)

Thus, the angular momentum of the electron in the 2^nd orbit is (h / π). Hence, the correct answer is (A) (h / π).