If `m,e,epsilon_(0) h` and `c` denote mass ,electron , change of electron, plank 's constant and speed of light , respectively , then the dimensions of `(me^(4))/(epsilon_(0)^(2) h^(2) c)` are

Options

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

Explanation

To find the dimensions of the expression \(\frac{m e^4}{\epsilon_0^2 h^2 c}\), we will analyze the dimensions of each variable involved.

### Step-by-Step Solution:

1. **Identify the dimensions of each variable:**
- Mass (\(m\)): The dimension of mass is \([M]\).
- Charge of electron (\(e\)): The dimension of electric charge is \([I T]\) (where \(I\) is electric current and \(T\) is time).
- Permittivity of free space (\(\epsilon_0\)): The dimension of \(\epsilon_0\) can be derived from the equation \(F = \frac{1}{4\pi \epsilon_0} \frac{q_1 q_2}{r^2}\), which gives \([\epsilon_0] = \frac{[M][L]^3}{[T]^4[I]^2}\).
- Planck's constant (\(h\)): The dimension of \(h\) is \([M][L]^2[T]^{-1}\).
- Speed of light (\(c\)): The dimension of speed is \([L][T]^{-1}\).

2. **Substituting the dimensions into the expression:**
\[ \text{Dimensions of } \frac{m e^4}{\epsilon_0^2 h^2 c} = \frac{[M] \cdot ([I T]^4)}{([\epsilon_0]^2) \cdot ([h]^2) \cdot ([c])} \]

3. **Calculating the dimensions of the denominator:**
- For \(\epsilon_0^2\):
\[ [\epsilon_0^2] = \left(\frac{[M][L]^3}{[T]^4[I]^2}\right)^2 = \frac{[M]^2[L]^6}{[T]^8[I]^4} \]
- For \(h^2\):
\[ [h^2] = ([M][L]^2[T]^{-1})^2 = [M]^2[L]^4[T]^{-2} \]
- For \(c\):
\[ [c] = [L][T]^{-1} \]

4. **Combining the dimensions of the denominator:**
\[ [\epsilon_0^2 h^2 c] = \frac{[M]^2[L]^6}{[T]^8[I]^4} \cdot [M]^2[L]^4[T]^{-2} \cdot [L][T]^{-1} \]
\[ = [M]^4[L]^{6+4+1}[T]^{-8-2-1}[I]^{-4} = [M]^4[L]^{11}[T]^{-11}[I]^{-4} \]

5. **Final expression for dimensions:**
Now substituting back into our original expression:
\[ \text{Dimensions of } \frac{m e^4}{\epsilon_0^2 h^2 c} = \frac{[M][I]^4[T]^4}{[M]^4[L]^{11}[T]^{-11}[I]^{-4}} \]
\[ = [M]^{1-4}[L]^{-11}[T]^{4+11}[I]^{4+4} = [M]^{-3}[L]^{-11}[T]^{15}[I]^{8} \]

### Final Result:
The dimensions of \(\frac{m e^4}{\epsilon_0^2 h^2 c}\) are:
\[ [M]^{-3}[L]^{-11}[T]^{15}[I]^{8} \]