The final speed of an electron accelerated from rest through a voltage of $V$ can be calculated using the principle of energy conservation, specifically the relationship between electric potential energy and kinetic energy.

When an electron is accelerated through a voltage $V$, its kinetic energy becomes equal to the electric potential energy:

$eV = \frac{1}{2} mv^2$

Where:

• $e = 1.602 \times 10^{-19} \, \text{C}$ (the charge of an electron),
• $m = 9.109 \times 10^{-31} \, \text{kg}$ (the mass of an electron),
• $V = 50 \times 10^3 \, \text{V}$ (the voltage through which the electron is accelerated),
• $v$ is the final speed of the electron.

Now, we can solve for $v$ (the final speed of the electron):

$v = \sqrt{\frac{2eV}{m}}$

Substituting the known values:

$v = \sqrt{\frac{2 \times (1.602 \times 10^{-19} \, \text{C}) \times (50 \times 10^3 \, \text{V})}{9.109 \times 10^{-31} \, \text{kg}}}$

Let me calculate that.It seems there was a timeout error during the calculation. I'll guide you through solving it manually:

The equation we need to compute is:

$v = \sqrt{\frac{2 \times (1.602 \times 10^{-19}) \times (50 \times 10^3)}{9.109 \times 10^{-31}}}$

Step-by-step:

1. Multiply the constants in the numerator: $2 \times (1.602 \times 10^{-19}) \times 50 \times 10^3 = 1.602 \times 10^{-14} \, \text{J}$

2. Divide by the electron mass: $\frac{1.602 \times 10^{-14}}{9.109 \times 10^{-31}} \approx 1.759 \times 10^{16}$

3. Take the square root: $v = \sqrt{1.759 \times 10^{16}} \approx 4.2 \times 10^8 \, \text{m/s}$

Thus, the final speed of the electron is approximately $4.2 \times 10^8 \, \text{m/s}$.

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Related Questions:

1. How does increasing the voltage affect the speed of an electron?
2. What would happen if the electron were replaced with a proton under the same conditions?
3. How does the mass of a particle influence its acceleration through a given potential?
4. What is the relationship between energy and voltage in this context?
5. How does relativistic speed correction apply if the speed approaches the speed of light?

Tip: When calculating the speed of particles near light speed, consider relativistic effects for greater accuracy.