Electrons are caused to fall through a potential difference of 1500 V. If they are initially at rest, their final speed is

Options

A. `23xx10^(7)m//s`
B. `2.3xx10^(7)m//s`
C. `32xx10^(7)m//s`
D. `3.2xx10^(7)m//s`

Explanation

To find the final speed of an electron that falls through a potential difference of 1500 V, we can use the principle of energy conservation. The kinetic energy gained by the electron is equal to the work done on it by the electric field, which can be expressed in terms of the charge and the potential difference.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Potential difference (V) = 1500 V
- Charge of an electron (Q) = \(1.6 \times 10^{-19}\) C
- Mass of an electron (m) = \(9.1 \times 10^{-31}\) kg

2. **Use the Energy Conservation Principle:**
The kinetic energy (KE) gained by the electron when it falls through a potential difference is given by:
\[ KE = Q \times V \]
Since the electron starts from rest, its initial kinetic energy is 0. Thus, the final kinetic energy is:
\[ KE = \frac{1}{2} m v^2 \]

3. **Set the Equations Equal:**
Equating the two expressions for kinetic energy, we have:
\[ \frac{1}{2} m v^2 = Q \times V \]

4. **Rearrange to Solve for Final Speed (v):**
\[ v^2 = \frac{2Q \times V}{m} \]
\[ v = \sqrt{\frac{2Q \times V}{m}} \]

5. **Substitute the Known Values:**
\[ v = \sqrt{\frac{2 \times (1.6 \times 10^{-19} \, \text{C}) \times (1500 \, \text{V})}{9.1 \times 10^{-31} \, \text{kg}}} \]

6. **Calculate the Value:**
- First, calculate the numerator:
\[ 2 \times (1.6 \times 10^{-19}) \times (1500) = 4.8 \times 10^{-16} \, \text{J} \]
- Now divide by the mass of the electron:
\[ \frac{4.8 \times 10^{-16}}{9.1 \times 10^{-31}} \approx 5.27 \times 10^{14} \]
- Finally, take the square root:
\[ v \approx \sqrt{5.27 \times 10^{14}} \approx 2.3 \times 10^7 \, \text{m/s} \]

### Final Answer:
The final speed of the electron is approximately \(2.3 \times 10^7 \, \text{m/s}\).

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