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

Options

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

Answer

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

### Step 1: Understand the quantity \( h \)
Planck's constant \( h \) is a fundamental physical constant that relates the energy of a photon to the frequency of its electromagnetic wave. The dimensions of \( h \) can be derived from the equation:
\[ E = h \cdot f \]
where \( E \) is energy and \( f \) is frequency.

### Step 2: Determine the dimensions of energy and frequency
The dimension of energy \( E \) is given by:
\[ [E] = [M][L^2][T^{-2}] \]
where \( M \) is mass, \( L \) is length, and \( T \) is time.

The dimension of frequency \( f \) is:
\[ [f] = [T^{-1}] \]

### Step 3: Find the dimensions of \( h \)
From the equation \( E = h \cdot f \), we can rearrange it to find the dimensions of \( h \):
\[ [h] = \frac{[E]}{[f]} = \frac{[M][L^2][T^{-2}]}{[T^{-1}]} = [M][L^2][T^{-2}] \cdot [T] = [M][L^2][T^{-1}] \]

### Step 4: Consider the factor \( 2\pi \)
The factor \( 2\pi \) is a dimensionless constant. Therefore, it does not affect the dimensions of \( h \).

### Step 5: Determine the dimensions of \( \frac{h}{2\pi} \)
Since \( 2\pi \) is dimensionless, the dimensions of \( \frac{h}{2\pi} \) are the same as the dimensions of \( h \):
\[ \left[\frac{h}{2\pi}\right] = [h] = [M][L^2][T^{-1}] \]

### Conclusion
Thus, the dimension of \( \frac{h}{2\pi} \) is:
\[ [M][L^2][T^{-1}] \]