The dimensions of `e^(2)//4pi epsilon_(0)hc`, where `e, epsilon_(0),h` and `c` are electronic charge, electric permittivity, Planck’s constant and velocity of light in vacuum respectively

Options

A. `[M^0L^0T^0]`
B. `[ML^0T^0]`
C. `[M^0LT^0]`
D. `[M^0L^0T^1]`

Answer

To find the dimensions of the expression \( \frac{e^2}{4 \pi \epsilon_0 h c} \), where \( e \) is the electronic charge, \( \epsilon_0 \) is the electric permittivity, \( h \) is Planck's constant, and \( c \) is the speed of light in vacuum, we will analyze the dimensions of each component step by step.

### Step 1: Determine the dimensions of \( e^2 \)

The dimension of electric charge \( e \) is given as:
\[ [e] = [I][T] = A \cdot T \]
Thus, the dimension of \( e^2 \) is:
\[ [e^2] = [I^2][T^2] = A^2 \cdot T^2 \]

### Step 2: Determine the dimensions of \( \epsilon_0 \)

The dimension of electric permittivity \( \epsilon_0 \) is given by:
\[ [\epsilon_0] = \frac{[M^{-1}][L^{-3}][T^4]}{[I^2]} = [M^{-1}][L^{-3}][T^4][I^{-2}] \]

### Step 3: Determine the dimensions of \( h \)

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

### Step 4: Determine the dimensions of \( c \)

The dimension of the speed of light \( c \) is given as:
\[ [c] = [L][T^{-1}] \]

### Step 5: Combine the dimensions in the expression

Now we will substitute these dimensions into the expression \( \frac{e^2}{4 \pi \epsilon_0 h c} \):

1. The numerator \( e^2 \):
\[ [e^2] = A^2 \cdot T^2 \]

2. The denominator \( 4 \pi \epsilon_0 h c \):
\[ [\epsilon_0] = [M^{-1}][L^{-3}][T^4][I^{-2}] \]
\[ [h] = [M][L^2][T^{-1}] \]
\[ [c] = [L][T^{-1}] \]

Therefore, the dimension of the denominator is:
\[ [\epsilon_0 h c] = [M^{-1}][L^{-3}][T^4][I^{-2}] \cdot [M][L^2][T^{-1}] \cdot [L][T^{-1}] \]

Simplifying this:
\[ = [M^{-1} \cdot M][L^{-3} \cdot L^2 \cdot L][T^4 \cdot T^{-1} \cdot T^{-1}][I^{-2}] \]
\[ = [L^{-3 + 2 + 1}][T^{4 - 1 - 1}][I^{-2}] \]
\[ = [L^0][T^2][I^{-2}] = [T^2][I^{-2}] \]

### Step 6: Final dimension of the expression

Now we can find the dimension of the entire expression:
\[ \frac{e^2}{4 \pi \epsilon_0 h c} = \frac{A^2 \cdot T^2}{T^2 \cdot I^{-2}} = A^2 \cdot I^2 \]

Since \( A = [I] \), we can express this as:
\[ = [I^2] \cdot [I^2] = [I^4] \]

### Conclusion

The final dimension of \( \frac{e^2}{4 \pi \epsilon_0 h c} \) is:
\[ [M^0][L^0][T^0] = 1 \quad \text{(dimensionless)} \]