Obtain the enrgy of joule acquired by an electron beam when accelerated through a potential difference of 2000 V. How much speed will the electron gain ?

Options

A. `(8)/(3)xx10^(7) m//s`
B. `(7)/(3)xx10^(7) m//s`
C. `(5)/(3)xx10^(7) m//s`
D. `(2)/(3) xx 10^(7) m//s`

Answer

To solve the problem, we need to find the energy acquired by an electron beam when it is accelerated through a potential difference of 2000 V, and then calculate the speed gained by the electron.

### Step 1: Calculate the energy acquired by the electron

The energy (in joules) acquired by an electron when it is accelerated through a potential difference \( V \) is given by the formula:

\[ E = e \cdot V \]

Where:
- \( E \) is the energy in joules,
- \( e \) is the charge of the electron, approximately \( 1.6 \times 10^{-19} \) coulombs,
- \( V \) is the potential difference in volts (2000 V in this case).

Substituting the values:

\[ E = (1.6 \times 10^{-19} \, \text{C}) \cdot (2000 \, \text{V}) \]

Calculating this gives:

\[ E = 3.2 \times 10^{-16} \, \text{J} \]

### Step 2: Calculate the speed gained by the electron

The kinetic energy acquired by the electron can also be expressed in terms of its mass and speed:

\[ E = \frac{1}{2} m v^2 \]

Where:
- \( m \) is the mass of the electron, approximately \( 9.1 \times 10^{-31} \, \text{kg} \),
- \( v \) is the speed of the electron.

Setting the two expressions for energy equal to each other:

\[ 3.2 \times 10^{-16} = \frac{1}{2} (9.1 \times 10^{-31}) v^2 \]

Rearranging to solve for \( v^2 \):

\[ v^2 = \frac{2 \cdot 3.2 \times 10^{-16}}{9.1 \times 10^{-31}} \]

Calculating the right side:

\[ v^2 = \frac{6.4 \times 10^{-16}}{9.1 \times 10^{-31}} \approx 7.02 \times 10^{14} \]

Taking the square root to find \( v \):

\[ v = \sqrt{7.02 \times 10^{14}} \approx 8.38 \times 10^7 \, \text{m/s} \]

### Final Answers:
- The energy acquired by the electron beam is \( 3.2 \times 10^{-16} \, \text{J} \).
- The speed gained by the electron is approximately \( 8.38 \times 10^7 \, \text{m/s} \).