An electron is released with a velocity of `5xx10^(6)ms^(-1)` in an electric field of `10^(3)NC^(-1)` which has been applied so as to oppose its motion. What distance would the electron travel and how much time could it take before it is brought to rest?

Options

A. `2.8xx10^(-8)s`
B. `1.8xx10^(-8)s`
C. `6.4xx10^(-8)s`
D. `4.2xx10^(-8)s`

Answer

a

Explanation

Since electric field is applied so as to oppose the motion of electron,
`a= -(eE)/(m)` (retardation)
`= -(1.6xx10^(-19)xx10^(3))/(9.1xx10^(-31))= 1.758xx10^(4)ms^(-2)`
Now, `u= 5xx10^(6) ms^(-1), v=0`
Using the relation: `v^(2)-u^(2)= 2`, as we get
`S= 7.11xx10^(-2)m`
Using the relation: `v=u+at`, we get,
`t= 2.844xx10^(-8)s`