Question

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

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

Answer 1

The Correct Option is A

To determine the momentum of an electron revolving in the \(n^{\text{th}}\) orbit, we need to apply the principles of quantum mechanics, particularly Bohr's model of the atom.

According to Bohr's model:

1. The angular momentum of an electron in the \(n^{\text{th}}\) orbit is quantized and given by the formula: L = n\hbar, where \hbar = \frac{h}{2\pi} and h is Planck's constant.
2. The angular momentum is also expressed as L = mvr, where m is the mass of the electron, v is its velocity, and r is the radius of the orbit.

By equating these two expressions for angular momentum:

mvr = n \frac{h}{2\pi}

To find the momentum p of the electron, which is defined as p = mv, we rearrange the relation above:

mv = \frac{nh}{2\pi r}

Thus, the formula for the momentum of the electron in the \(n^{\text{th}}\) orbit is:

p = \frac{nh}{2\pi r}

This matches the given correct answer option: \(\frac{ nh }{2 \pi r }\).

Therefore, based on the given options and Bohr's model, the momentum of an electron in the \(n^{\text{th}}\) orbit is correctly represented by the first option \(\frac{ nh }{2 \pi r }\).