In an electron gun, the potential difference between the filament and plate is 3000 V. What will be the velocity of electron emitting from the gun?

Options

A. `3xx10^(8) m//s`
B. `3.18xx10^(7) m//s`
C. `3.52xx10^(7) m//s`
D. `3.26xx10^(7) m//s`

Answer

To find the velocity of an electron emitted from an electron gun with a potential difference of 3000 V, we can use the principle of energy conservation. The kinetic energy gained by the electron when it is accelerated through the potential difference is equal to the work done on it by the electric field.

### Step-by-Step Solution:

1. **Understand the relationship between potential difference and kinetic energy**:
The work done on the electron by the electric field when it is accelerated through a potential difference \( V \) is given by:
\[ W = eV \]
where \( e \) is the charge of the electron (approximately \( 1.6 \times 10^{-19} \) coulombs) and \( V \) is the potential difference.

2. **Set the work done equal to the kinetic energy**:
The kinetic energy (KE) of the electron when it is emitted can be expressed as:
\[ KE = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the electron (approximately \( 9.11 \times 10^{-31} \) kg) and \( v \) is the velocity of the electron.

3. **Equate the two expressions**:
Setting the work done equal to the kinetic energy gives us:
\[ eV = \frac{1}{2} mv^2 \]

4. **Rearranging the equation to solve for velocity**:
Rearranging the equation to solve for \( v \):
\[ v^2 = \frac{2eV}{m} \]
\[ v = \sqrt{\frac{2eV}{m}} \]

5. **Substituting the known values**:
Now, substitute \( e = 1.6 \times 10^{-19} \) C, \( V = 3000 \) V, and \( m = 9.11 \times 10^{-31} \) kg into the equation:
\[ v = \sqrt{\frac{2 \times (1.6 \times 10^{-19}) \times 3000}{9.11 \times 10^{-31}}} \]

6. **Calculating the value**:
First calculate the numerator:
\[ 2 \times (1.6 \times 10^{-19}) \times 3000 = 9.6 \times 10^{-16} \]
Now divide by the mass of the electron:
\[ \frac{9.6 \times 10^{-16}}{9.11 \times 10^{-31}} \approx 1.055 \times 10^{15} \]
Now take the square root:
\[ v \approx \sqrt{1.055 \times 10^{15}} \approx 1.027 \times 10^{7} \text{ m/s} \]

### Final Answer:
The velocity of the electron emitting from the gun is approximately \( 1.027 \times 10^{7} \) m/s.