Question

An electron is moving in a straight line with a velocity of 4.0×105 m/s. It enters a region 5.0 cm long where it undergoes an acceleration of 6.0×1012m/s2 along the same straight line. (a) What is the electron’s velocity when it emerges from this region? b) How long does the electron take to cross the region?

Answer 1

(a) $8.7\cdot 10^5 m/s$

We can solve this part of the problem by using the following SUVAT equation:

$v^2-u^2=2ad$

where

v is the final velocity

u is the initial velocity

a is the acceleration

d is the distance through which the electron is accelerated

In this problem,

$u = 4.0\cdot 10^5 m/s\\a = 6.0\cdot 10^{12} m/s^2\\d = 5.0 cm = 0.05 m$

Solving for v,

$v=\sqrt{u^2+2ad}=\sqrt{(4.0\cdot 10^5)^2+2(6.0\cdot 10^{12})(0.05)}=8.7\cdot 10^5 m/s$

(b) $7.9\cdot 10^{-8}s$

The time needed for the electron to cross the region where it is accelerated can be found by using the following SUVAT equation:

$d=(\frac{v+u}{2})t$

where we have:

d = 5 cm = 0.05 m

$v=8.7\cdot 10^5 m/s$

$u=4.0\cdot 10^5 m/s$

Solving for t, we find:

$t=\frac{2d}{u+v}=\frac{2(0.05)}{8.7\cdot 10^5+4.0\cdot 10^5}=7.9\cdot 10^{-8}s$