To calculate the final speed of an electron accelerated from rest through a voltage of 50 kV, we use the principle of energy conservation. Specifically, the electrical potential energy gained by the electron is converted into kinetic energy.

The formula relating the kinetic energy to the electric potential is:

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

Where:

• $e = 1.602 \times 10^{-19} \, \text{C}$ is the charge of the electron,
• $V = 50 \, \text{kV} = 50 \times 10^3 \, \text{V}$ is the potential difference,
• $m = 9.109 \times 10^{-31} \, \text{kg}$ is the mass of the electron,
• $v$ is the final speed of the electron.

Step 1: Rearrange the equation to solve for $v$:

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

Step 2: Plug in the values:

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

I'll calculate the result.The final speed of the electron after being accelerated through a voltage of 50 kV is approximately $1.33 \times 10^8 \, \text{m/s}$.

Would you like more details or have any further questions?

Here are some related questions:

1. What is the speed of an electron when accelerated by a 100 kV potential?
2. How does the mass of a particle affect its final speed when accelerated through a voltage?
3. How would this calculation change if we were dealing with a proton instead of an electron?
4. What is the concept of relativistic effects at high speeds?
5. How does the potential energy of a charged particle vary with voltage?

Tip: As particle speeds approach the speed of light, relativistic effects must be considered for greater accuracy.