`h/(2pi)` is the dimension of

Options

A. Velocity
B. Momentum
C. Energy
D. Angular momentum

Answer

To find the dimensions of \( \frac{h}{2\pi} \), we can follow these steps:

### Step 1: Understand the context
We know from Bohr's postulate that the angular momentum \( L \) of an electron in an orbit can be expressed as an integral multiple of \( \frac{h}{2\pi} \). Therefore, we need to find the dimensions of angular momentum.

### Step 2: Write the formula for angular momentum
Angular momentum \( L \) is given by the formula:
\[ L = M \cdot v \cdot r \]
where:
- \( M \) is the mass of the electron,
- \( v \) is the velocity,
- \( r \) is the radius of the orbit.

### Step 3: Determine the dimensions of each component
1. **Mass (M)**: The dimension of mass is \( [M] \).
2. **Velocity (v)**: Velocity is defined as distance per unit time, so its dimension is:
\[ [v] = \frac{[L]}{[T]} = LT^{-1} \]
3. **Radius (r)**: The dimension of radius is simply:
\[ [r] = [L] \]

### Step 4: Combine the dimensions
Now, substituting the dimensions into the angular momentum formula:
\[ [L] = [M] \cdot [v] \cdot [r] = [M] \cdot (LT^{-1}) \cdot [L] \]
This simplifies to:
\[ [L] = [M] \cdot [L]^2 \cdot [T]^{-1} \]
Thus, the dimensions of angular momentum are:
\[ [L] = M L^2 T^{-1} \]

### Step 5: Relate this to \( \frac{h}{2\pi} \)
Since \( L \) is equal to \( n \cdot \frac{h}{2\pi} \) (where \( n \) is an integer), we can conclude that:
\[ \frac{h}{2\pi} \text{ has the same dimensions as } L \]
Therefore, the dimensions of \( \frac{h}{2\pi} \) are:
\[ \frac{h}{2\pi} \sim M L^2 T^{-1} \]

### Final Answer
The dimension of \( \frac{h}{2\pi} \) is \( M L^2 T^{-1} \).

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