Question

The momentum of an electron revolving in n^th orbit is given by: (Symbols have their usual meanings)

• A. \(\frac{nh}{2πr}\)
• B. \(\frac{nh}{2r}\)
• C. \(\frac{nh}{2\pi}\)
• D. \(\frac{2πr}{nh}\)

Answer 1

The Correct Option is A

To find the expression for the momentum of an electron revolving in the n^th orbit, we start by understanding the basic principles of atomic structure, particularly focusing on Bohr's model of the atom.

According to Bohr's postulate for quantized angular momentum, the angular momentum (L) of an electron in the n^th orbit is given by:

L = n\hbar

where \hbar is the reduced Planck's constant: \hbar = \frac{h}{2\pi}, and n is the principal quantum number.

Since the angular momentum (L) is also expressed as the product of linear momentum (p) and radius (r) of the orbit, we have:

L = p \cdot r

Equating the two expressions for angular momentum, we get:

p \cdot r = n\hbar

Substituting the value of \hbar, we get:

p \cdot r = n \cdot \frac{h}{2\pi}

Solving for the linear momentum (p):

p = \frac{n \cdot h}{2\pi r}

This shows that the momentum of an electron revolving in the n^th orbit is given by: \(\frac{nh}{2\pi r}\), which matches the correct answer option.

Conclusion: The correct expression for the momentum of an electron in the n^th orbit is \(\frac{nh}{2\pi r}\). This expression adheres to the Bohr model by quantifying the orbit characteristics and aligning with the angular momentum quantization rule.