The dimension of `e^(2)//epsilon_(0)hc` (here symbols have their usual meanings) are

Options

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

Answer

To find the dimensions of the expression \( \frac{e^2}{\epsilon_0 hc} \), we will analyze the dimensions of each component in the expression step by step.

### Step 1: Identify the dimensions of each component

1. **Charge \( e \)**:
The dimension of electric charge is given by:
\[ [e] = [I][T] \]
where \( [I] \) is the dimension of current and \( [T] \) is the dimension of time.

2. **Permittivity of free space \( \epsilon_0 \)**:
The dimension of permittivity \( \epsilon_0 \) can be derived from the equation of capacitance:
\[ C = \frac{Q}{V} \quad \text{and} \quad V = \frac{F}{Q} \quad \Rightarrow \quad \epsilon_0 = \frac{Q^2}{F \cdot L^2} \]
The dimension of force \( F \) is \( [M][L][T^{-2}] \), and the dimension of voltage \( V \) is \( [M][L^2][T^{-3}][I^{-1}] \). Thus, we can express:
\[ [\epsilon_0] = \frac{[I^2][T^2]}{[M][L][T^{-2}]} = \frac{[I^2][T^2]}{[M][L]} = [M^{-1}][L^{-3}][T^4][I^2] \]

3. **Planck's constant \( h \)**:
The dimension of Planck's constant is:
\[ [h] = [M][L^2][T^{-1}] \]

4. **Speed of light \( c \)**:
The dimension of speed is:
\[ [c] = [L][T^{-1}] \]

### Step 2: Substitute the dimensions into the expression

Now we can substitute these dimensions into the expression \( \frac{e^2}{\epsilon_0 hc} \):

1. **Dimension of \( e^2 \)**:
\[ [e^2] = [I^2][T^2] \]

2. **Dimension of \( \epsilon_0 \)**:
\[ [\epsilon_0] = [M^{-1}][L^{-3}][T^4][I^2] \]

3. **Dimension of \( h \)**:
\[ [h] = [M][L^2][T^{-1}] \]

4. **Dimension of \( c \)**:
\[ [c] = [L][T^{-1}] \]

### Step 3: Combine the dimensions

Now we can combine these dimensions in the expression:
\[ \frac{[e^2]}{[\epsilon_0][h][c]} = \frac{[I^2][T^2]}{[M^{-1}][L^{-3}][T^4][I^2] \cdot [M][L^2][T^{-1}] \cdot [L][T^{-1}]} \]

### Step 4: Simplify the expression

Now we simplify the denominator:
\[ [\epsilon_0][h][c] = [M^{-1}][L^{-3}][T^4][I^2] \cdot [M][L^2][T^{-1}] \cdot [L][T^{-1}] = [L^{-3}][T^4][I^2][L^2][T^{-1}][L][T^{-1}] \]
\[ = [L^{-3 + 2 + 1}][T^{4 - 1 - 1}][I^2] = [L^{-0}][T^{2}][I^2] = [T^2][I^2] \]

Now, substituting back:
\[ \frac{[I^2][T^2]}{[T^2][I^2]} = 1 \]

### Conclusion

The dimensions of \( \frac{e^2}{\epsilon_0 hc} \) are dimensionless, which can be represented as:
\[ [M^0][L^0][T^0] \]

### Final Answer
The dimensions of \( \frac{e^2}{\epsilon_0 hc} \) are \( M^0 L^0 T^0 \) (dimensionless).

---