The work required to stop the electron is equal to the kinetic energy of the electron since work is needed to reduce the electron's velocity to zero. The formula for kinetic energy ($KE$) is:

$KE = \frac{1}{2} m v^2$

Given:

• Mass of the electron ($m$) = $9.11 \times 10^{-31} \, \text{kg}$
• Velocity ($v$) = $1.90 \times 10^6 \, \text{m/s}$

Calculation:

$KE = \frac{1}{2} (9.11 \times 10^{-31}) (1.90 \times 10^6)^2$

1. First, calculate $v^2$: $v^2 = (1.90 \times 10^6)^2 = 3.61 \times 10^{12}$

2. Substitute values into the kinetic energy formula: $KE = \frac{1}{2} \times 9.11 \times 10^{-31} \times 3.61 \times 10^{12}$

3. Simplify: $KE = 1.645 \times 10^{-18} \, \text{J}$

Final Answer:

The work required to stop the electron is $1.645 \times 10^{-18} \, \text{J}$.

---

Would you like more details or have further questions? Here are related topics to explore:

1. How is the concept of work-energy theorem applied in other systems?
2. What is the significance of electron mass in quantum mechanics?
3. How would the result change if the velocity doubled?
4. What are practical examples of stopping an electron in physics experiments?
5. How do relativistic effects modify kinetic energy at high speeds?

Tip: Always check units to ensure consistency in calculations!