If h is the Planck's constant, m = mass of the electron, e = charge of the electron and `in_(0)` = permittivity of vacuum, then `(h^(2)in_(0))/(me^(2))` has the unit

Options

A. newton
B. joule
C. watt
D. metre

Answer

To find the units of the expression \((\frac{h^2 \epsilon_0}{m e^2})\), we will break down the units of each component step by step.

### Step 1: Identify the units of each component
1. **Planck's constant (h)**: The dimensions of Planck's constant are given as:
\[ [h] = M^1 L^2 T^{-1} \]
where \(M\) is mass, \(L\) is length, and \(T\) is time.

2. **Permittivity of vacuum (\(\epsilon_0\))**: The dimensions of permittivity of vacuum are:
\[ [\epsilon_0] = M^{-1} L^{-3} T^4 I^2 \]

3. **Mass of the electron (m)**: The dimensions of mass are:
\[ [m] = M^1 \]

4. **Charge of the electron (e)**: The dimensions of charge are:
\[ [e] = I^1 T^1 \]

### Step 2: Substitute the units into the expression
We need to substitute these dimensions into the expression \((\frac{h^2 \epsilon_0}{m e^2})\).

\[ [h^2] = (M^1 L^2 T^{-1})^2 = M^2 L^4 T^{-2} \]

\[ [e^2] = (I^1 T^1)^2 = I^2 T^2 \]

### Step 3: Combine the units in the expression
Now substituting these into our expression:

\[ \frac{h^2 \epsilon_0}{m e^2} = \frac{(M^2 L^4 T^{-2})(M^{-1} L^{-3} T^4 I^2)}{M^1 (I^2 T^2)} \]

### Step 4: Simplify the expression
Now we simplify the expression:

1. **Numerator**:
\[ M^2 L^4 T^{-2} \cdot M^{-1} L^{-3} T^4 I^2 = M^{2-1} L^{4-3} T^{-2+4} I^2 = M^1 L^1 T^2 I^2 \]

2. **Denominator**:
\[ M^1 (I^2 T^2) = M^1 I^2 T^2 \]

Now, putting it all together:

\[ \frac{M^1 L^1 T^2 I^2}{M^1 I^2 T^2} \]

### Step 5: Cancel out the common terms
When we cancel out the common terms \(M^1\), \(I^2\), and \(T^2\):

\[ = L^1 \]

### Final Result
Thus, the dimensions of the expression \((\frac{h^2 \epsilon_0}{m e^2})\) simplify to:

\[ L^1 \]

This means the unit of the expression is **meters (m)**.