The dimension of the quantity `1/epsilon_0 e^2/(hc)` is (e charge of electron,h Planck's constant and c=velocity of light)

Options

A. `[M^(-1) L^(-3) T^2 A]`
B. `[M^(0) L^(0) T^0 A^0]`
C. `[ML^(3) T^(-4) A^(-2)]`
D. `[M^(-1) L^(-3) T^4 A^2]`

Answer

To find the dimension of the quantity \( \frac{1}{\epsilon_0} \frac{e^2}{hc} \), we will analyze the dimensions of each component involved.

### Step 1: Identify the dimensions of \( \epsilon_0 \)
The permittivity of free space \( \epsilon_0 \) has the dimension given by:
\[ [\epsilon_0] = \frac{1}{\text{Force}} \cdot \frac{\text{Length}^2}{\text{Charge}^2} = \frac{1}{M L T^{-2}} \cdot L^2 A^{-2} \]
This simplifies to:
\[ [\epsilon_0] = M^{-1} L^{-3} T^4 A^2 \]

### Step 2: Identify the dimensions of \( e^2 \)
The charge \( e \) has the dimension of electric charge:
\[ [e] = A \cdot T \]
Thus, the dimension of \( e^2 \) is:
\[ [e^2] = A^2 T^2 \]

### Step 3: Identify the dimensions of \( h \)
Planck's constant \( h \) has the dimension of action, which is:
\[ [h] = M L^2 T^{-1} \]

### Step 4: Identify the dimensions of \( c \)
The speed of light \( c \) has the dimension of velocity:
\[ [c] = L T^{-1} \]

### Step 5: Combine the dimensions in the expression \( \frac{e^2}{hc} \)
Now we can find the dimensions of \( hc \):
\[ [hc] = [h][c] = (M L^2 T^{-1})(L T^{-1}) = M L^3 T^{-2} \]
Now, we can find the dimensions of \( \frac{e^2}{hc} \):
\[ \left[\frac{e^2}{hc}\right] = \frac{[e^2]}{[hc]} = \frac{A^2 T^2}{M L^3 T^{-2}} = \frac{A^2 T^4}{M L^3} \]

### Step 6: Find the dimensions of \( \frac{1}{\epsilon_0} \)
Now, we take the reciprocal of \( \epsilon_0 \):
\[ \left[\frac{1}{\epsilon_0}\right] = M L^3 T^{-4} A^{-2} \]

### Step 7: Combine the dimensions of \( \frac{1}{\epsilon_0} \) and \( \frac{e^2}{hc} \)
Now we can find the dimensions of the entire expression:
\[ \left[\frac{1}{\epsilon_0} \frac{e^2}{hc}\right] = [\frac{1}{\epsilon_0}] \cdot [\frac{e^2}{hc}] = (M L^3 T^{-4} A^{-2}) \cdot \left(\frac{A^2 T^4}{M L^3}\right) \]
When we multiply these, we get:
\[ = M^0 L^0 T^0 A^0 = 1 \]

### Conclusion
Thus, the dimension of the quantity \( \frac{1}{\epsilon_0} \frac{e^2}{hc} \) is dimensionless (or \( M^0 L^0 T^0 A^0 \)).