Chapter 6, Problem 17P

How much work is required to stop an electron $(m=9.11\times {10}^{-31}kg)$ which is moving with a speed of $1.10\times {10}^{6}m/s$ ?

To determine

Work required to stop an electron.

Answer to Problem 17P

Solution: $5.51\times {10}^{-19}\text{J}$

Explanation of Solution

Given:Mass of electron, $m=9.11\times {10}^{-31}\text{kg}$

Velocity of electron, $v=1.10\times {10}^{\text{6}}\text{m/s}$

Formula Used:

According to work-energy theorem, work done by the force equal to change in kinetic energy of the particle is:

  $\begin{array}{l}W=\Delta K.E\\ W=K.E_{final}-K.E_{initial}\end{array}$

Here final kinetic energy is zero

  $W=-K.E_{initial}$

  $W=-\frac{1}{2}m{v}^2\mathrm{.........}(1)$

Calculations:

Substitute this value in equation(1)

  $\begin{array}{l}$ W=-\frac{1}{2}9.11\times {10}^{-31}\times {(1.10\times {10}^6)}^2$W=-5.51\times {10}^{-19}\text{ J}\end{array}$

Work required to stop an electron is 5.51×10-19 J

Conclusion: $5.51\times {10}^{-19}\text{J}$ work required to stop an electron of mass $9.11\times {10}^{-31}\text{kg}$ which is moving at $1.10\times {10}^6\text{m/s}.$