The magnitude of electric field intensity E is such that, an electron placed in it would experience an electrical force equal to its weight is given by

Options

A. `mge`
B. `(mg)/(e )`
C. `(e )/(mg)`
D. `(e^(2) g)/( 2m)`

Answer

To solve the problem, we need to find the magnitude of the electric field intensity \( E \) such that the electric force experienced by an electron in that field is equal to its weight.

### Step-by-step Solution:

1. **Understand the forces acting on the electron**:
- The weight \( W \) of the electron can be expressed as:
\[ W = mg \]
where \( m \) is the mass of the electron and \( g \) is the acceleration due to gravity.

2. **Identify the electric force acting on the electron**:
- The electric force \( F \) experienced by a charge \( q \) in an electric field \( E \) is given by:
\[ F = qE \]
- For an electron, the charge \( q \) is equal to the elementary charge \( e \) (approximately \( 1.6 \times 10^{-19} \) coulombs).

3. **Set the weight equal to the electric force**:
- According to the problem, the electric force is equal to the weight of the electron:
\[ mg = eE \]

4. **Rearrange the equation to solve for the electric field \( E \)**:
- We can rearrange the equation to isolate \( E \):
\[ E = \frac{mg}{e} \]

5. **Final expression**:
- The magnitude of the electric field intensity \( E \) such that the electric force on the electron equals its weight is:
\[ E = \frac{mg}{e} \]

### Conclusion:
The correct answer is \( \frac{mg}{e} \).