Question

An electron initially at rest is accelerated over a distance of 0.210 m in 33.3 ns. Assuming its acceleration is constant, what voltage was used to accelerate it

Answer 1

V = 451.47 volts

Step-by-step explanation:

Given that,

Distance, d = 0.21 m

Initial speed, u = 0

Time, t = 33.3 ns

Let v is the final velocity. Using second equation of motion as :

`d=ut+(1)/(2)at^2`

a is acceleration,
`a=(v-u)/(t)` and u = 0

So,

`d=(1)/(2)(v-u)t`

`v=(2d)/(t)\\\\v=(2* 0.21)/(33.3* 10^(-9))\\\\v=1.26* 10^7\ m/s`

Now applying the conservation of energy i.e.

`(1)/(2)mv^2=qV`

V is voltage

`V=(mv^2)/(2q)\\\\V=(9.1* 10^(-31)* (1.26* 10^7)^2)/(2* 1.6* 10^(-19))\\\\V=451.47\ V`

So, the voltage is 451.47 V.