The bohr radius is given by `a_(0) = (epsilon_(0)h^(2))/(pi m e^(2))` verify that the RHS has dimesions of length

Explanation

To verify that the right-hand side (RHS) of the equation for the Bohr radius \( a_0 = \frac{\epsilon_0 h^2}{\pi m e^2} \) has dimensions of length, we will analyze the dimensions of each component in the equation step by step.

### Step 1: Identify the components and their dimensions

1. **Electric constant (\( \epsilon_0 \))**:
From Coulomb's law, we know that:
\[ F = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2} \]
Rearranging gives:
\[ \epsilon_0 = \frac{q^2}{F r^2} \]
The dimensions of charge \( q \) are \( [Q] = A \cdot T \) (Ampere-second), so:
\[ [q^2] = [A^2 T^2] \]
The dimensions of force \( F \) are \( [F] = M L T^{-2} \), and the dimensions of radius \( r \) are \( [L] \). Therefore:
\[ [\epsilon_0] = \frac{[A^2 T^2]}{[M L T^{-2}] [L^2]} = \frac{[A^2 T^2]}{[M L^3 T^{-2}]} = \frac{[A^2 T^4]}{[M L^3]} \]

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

3. **Mass (\( m \))**:
The dimensions of mass \( m \) are simply:
\[ [m] = [M] \]

4. **Charge (\( e \))**:
The dimensions of charge \( e \) are the same as \( q \):
\[ [e] = [A T] \]
Therefore:
\[ [e^2] = [A^2 T^2] \]

### Step 2: Substitute dimensions into the RHS

Now we substitute the dimensions into the RHS of the equation:
\[ a_0 = \frac{\epsilon_0 h^2}{\pi m e^2} \]

Ignoring the dimensionless constant \( \pi \), we have:
\[ [a_0] = \frac{[\epsilon_0] [h]^2}{[m] [e]^2} \]

Substituting the dimensions we found:
\[ [a_0] = \frac{\frac{[A^2 T^4]}{[M L^3]} \cdot ([M L^2 T^{-1}])^2}{[M] \cdot [A^2 T^2]} \]

### Step 3: Simplifying the expression

Now we simplify the expression:
1. The numerator becomes:
\[ \frac{[A^2 T^4]}{[M L^3]} \cdot [M^2 L^4 T^{-2}] = \frac{[A^2 T^4 M^2 L^4]}{[M L^3]} \]
This simplifies to:
\[ [A^2 T^4 M L] \quad \text{(after canceling one M and one L from denominator)} \]

2. The denominator is:
\[ [M] \cdot [A^2 T^2] = [M A^2 T^2] \]

Putting it all together:
\[ [a_0] = \frac{[A^2 T^4 M L]}{[M A^2 T^2]} = [L] \cdot [T^2] = [L] \]

### Conclusion

Thus, we find that the dimensions of \( a_0 \) indeed simplify to \( [L] \), confirming that the RHS has dimensions of length.