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

The electron's velocity when it emerges from the region is 4.3 × 10^10 m/s. It takes 7.1667 × 10^-6 seconds for the electron to cross the region.

Step-by-step explanation:

In order to find the electron's velocity when it emerges from the region, we can use the equation:

vf = vi + at

where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time taken to cross the region.

Given that the initial velocity is 4.0 × 105 m/s, the acceleration is 6.0 × 1012 m/s2, and the distance traveled is 5.0 cm = 0.05 m, we can solve for the final velocity:

vf = (4.0 × 105) + (6.0 × 1012) × (0.05)

vf = 4.3 × 1010 m/s

To find the time taken to cross the region, we can use the formula:

t = (vf - vi) / a

Substituting the given values:

t = (4.3 × 1010 - 4.0 × 105) / 6.0 × 1012

t = 7.1667 × 10-6 seconds

Answer 2

Step-by-step explanation:

Given that,

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

Distance, s = 5 cm = 0.05 cm

Acceleration of the electron,
`a=6* 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* 10^5)^2+2* 6* 10^(12)* 0.05`

v = 871779.788 m/s

or

`v=8.71* 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=(v-u)/(a)`

`t=(8.71* 10^5-4* 10^5)/(6* 10^(12))`

`t=7.85* 10^(-8)\ s`

Hence, this is the required solution.