If electric charge e, electron's mass m, speed of light in vaccum c and Planck's constant h are taken as fundamental quantities, the permeability of vaccum `mu_(0)` can be expressed in units of :

Options

A. `((hc)/(me^2))`
B. `((h)/(me^2))`
C. `((h)/(ce^2))`
D. `((mc^2)/(he^2))`

Explanation

To express the permeability of vacuum \( \mu_0 \) in terms of the fundamental quantities: electric charge \( e \), electron's mass \( m \), speed of light in vacuum \( c \), and Planck's constant \( h \), we can use dimensional analysis. Let's go through the steps systematically.

### Step 1: Identify the dimensions of the quantities involved
1. **Electric Charge \( e \)**: The dimension of electric charge is given by \( [e] = [I][T] \) (where \( I \) is current and \( T \) is time).
2. **Mass \( m \)**: The dimension of mass is \( [m] = [M] \).
3. **Speed of Light \( c \)**: The dimension of speed is \( [c] = [L][T]^{-1} \).
4. **Planck's Constant \( h \)**: The dimension of Planck's constant is \( [h] = [M][L]^2[T]^{-1} \).

### Step 2: Write the dimension of permeability of vacuum \( \mu_0 \)
The dimension of permeability of vacuum is given by:
\[ [\mu_0] = [M][L][T]^{-2}[A]^{-2} \]
where \( [A] \) is the dimension of electric current.

### Step 3: Express \( \mu_0 \) in terms of the fundamental quantities
We want to express \( \mu_0 \) in the form:
\[ \mu_0 = e^x m^y c^z h^w \]
where \( x, y, z, w \) are the powers we need to determine.

### Step 4: Set up the equations based on dimensions
From the dimensions, we have:
\[ [M][L][T]^{-2}[A]^{-2} = [I]^x [M]^y [L]^z [T]^{-2z} [M]^w [L]^{2w} [T]^{-w} \]

### Step 5: Equate the dimensions
Now we can equate the powers of each dimension:

1. For \( M \):
\[ 1 = y + w \]
2. For \( L \):
\[ 1 = z + 2w \]
3. For \( T \):
\[ -2 = -2z - w \]
4. For \( A \):
\[ -2 = x \]

### Step 6: Solve the equations
From the equation for \( A \):
\[ x = -2 \]

Substituting \( x = -2 \) into the equation for \( T \):
\[ -2 = -2z - w \implies w = -2 - 2z \]

Now substituting \( w \) into the equation for \( M \):
\[ 1 = y + (-2 - 2z) \implies y = 3 + 2z \]

Substituting \( w \) into the equation for \( L \):
\[ 1 = z + 2(-2 - 2z) \implies 1 = z - 4 - 4z \implies 5z = 5 \implies z = 1 \]

Now substituting \( z = 1 \) back to find \( w \):
\[ w = -2 - 2(1) = -4 \]

Now substituting \( z = 1 \) into \( y \):
\[ y = 3 + 2(1) = 5 \]

### Step 7: Write the final expression for \( \mu_0 \)
Now we have:
- \( x = -2 \)
- \( y = 5 \)
- \( z = 1 \)
- \( w = -4 \)

Thus, we can express \( \mu_0 \) as:
\[ \mu_0 = e^{-2} m^{5} c^{1} h^{-4} \]

### Conclusion
Rearranging gives us:
\[ \mu_0 = \frac{h}{e^2 c} \]

### Final Answer
The permeability of vacuum \( \mu_0 \) can be expressed in units of:
\[ \mu_0 = \frac{h}{e^2 c} \]