Question

89. 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

Explanation:

Given that,

Initial speed of the electron, $u=4\times 10^5\ m/s$

Distance, s = 5 cm = 0.05 cm

Acceleration of the electron, $a=6\times 10^{12}\ m/s^2$  

(a) Let v is the electron's velocity when it emerges from this region. It can be calculated as :

$v^2=u^2+2as$

$v^2=(4\times 10^5)^2+2\times 6\times 10^{12}\times 0.05$

v = 871779.788 m/s

or

$v=8.71\times 10^5\ m/s$

(b) Let t is the time for which the electron take to cross the region. It can be calculated as:

$t=\dfrac{v-u}{a}$

$t=\dfrac{8.71\times 10^5-4\times 10^5}{6\times 10^{12}}$

$t=7.85\times 10^{-8}\ s$

Hence, this is the required solution.