The parameter `(mQ^(4))/(epsilon_(0)^(2)h^(2))` has the dimensions of ( m = mass Q = charge `in_(0)` = Permittivity and h = Planck's constant )

Options

A. Wavelength
B. Power
C. Angular momentum
D. Energy

Answer

To find the dimensions of the parameter \(\frac{m Q^4}{\epsilon_0^2 h^2}\), we will break down the dimensions of each component involved: mass \(m\), charge \(Q\), permittivity \(\epsilon_0\), and Planck's constant \(h\).

### Step 1: Identify the dimensions of each parameter
1. **Mass \(m\)**:
- Dimension: \([M] = M^1 L^0 T^0\)

2. **Charge \(Q\)**:
- Dimension: \([Q] = A^1 T^1\) (where \(A\) is the dimension of electric current)

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

4. **Planck's constant \(h\)**:
- Dimension: \([h] = M^1 L^2 T^{-1}\)

### Step 2: Substitute the dimensions into the expression
Now, we substitute these dimensions into the expression \(\frac{m Q^4}{\epsilon_0^2 h^2}\).

- **Numerator**:
\[ m Q^4 = M^1 L^0 T^0 \cdot (A^1 T^1)^4 = M^1 L^0 T^0 \cdot A^4 T^4 = M^1 A^4 T^4 \]

- **Denominator**:
\[ \epsilon_0^2 = (M^{-1} L^{-3} T^4 A^2)^2 = M^{-2} L^{-6} T^8 A^4 \]
\[ h^2 = (M^1 L^2 T^{-1})^2 = M^2 L^4 T^{-2} \]

Now, combining the denominator:
\[ \epsilon_0^2 h^2 = (M^{-2} L^{-6} T^8 A^4) \cdot (M^2 L^4 T^{-2}) = M^{0} L^{-2} T^{6} A^{4} \]

### Step 3: Combine the numerator and denominator
Now we can combine the numerator and denominator:
\[ \frac{M^1 A^4 T^4}{M^{0} L^{-2} T^{6} A^{4}} = M^{1-0} A^{4-4} T^{4-6} L^{0-(-2)} = M^1 A^0 T^{-2} L^{2} \]

### Step 4: Final dimensions
Thus, the final dimensions of the parameter \(\frac{m Q^4}{\epsilon_0^2 h^2}\) are:
\[ [M^1 L^2 T^{-2}] \]

This is the dimension of energy.