Question

(1) How much work is required to stop an electron
(m = 9.11 X 10-31 kg) which is moving with a speed
of 1.90 x 10 m/s?

Answer 1

Answer:

$-1.64\cdot 10^{-18}J$

Explanation:

We can solve the problem by using the work-energy theorem: in fact, the work done on the electron must be equal to its change in kinetic energy, therefore

$W=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2$

where

$m=9.11\cdot 10^{-31} kg$ is the mass of the electron

v = 0 is the final velocity of the electron (it comes to a stop)

u is the initial velocity of the electron

In the problem, we are told that the electron is initially moving at

$u=1.90\cdot 10^6 m/s$

Therefore, the work required to stop is is

$W=-\frac{1}{2}mu^2=-\frac{1}{2}(9.11\cdot 10^{-31})(1.90\cdot 10^6)^2=-1.64\cdot 10^{-18}J$

and the work is negative since it is in the opposite direction to the motion of the electron.