A particle of mass `6.4xx10^(-27)` kg and charge `3.2xx10^(-19)C` is situated in a uniform electric field of `1.6xx10^5 Vm^(-1)`. The velocity of the particle at the end of `2xx10^(-2)`m path when it starts from rest is :

Options

A. `2sqrt(3) xx 10^(5) ms^(-1)`
B. `8 xx 10^(5) ms^(-1)`
C. `16 xx 10^(5) ms^(-1)`
D. `4sqrt(2) xx 10^(5) ms^(-1)`

Answer

D

Explanation

Here, `m_(a) = 6.4 xx 10^(-27) kg, q_(alpha) = 3.2 xx 10^(-19)C, E = 1.6 xx 10^(5) Vm^(-1), S = 2 xx 10^(-2)m`,
Force on `alpha`-particle in a uniform electric field is
`F = q_(alpha)E = (3.2 xx 10^(-19)) xx (1.6 xx 10^(5))N = 3.2 xx 1.6 xx 10^(-14) N`
Acceleration of the `alpha`-particle is
`a = (F)/(m_(alpha)) = (3.2 xx 1.6 xx 10^(-14))/(6.4 xx 10^(-27))ms^(-2) = 8 xx 10^(12) ms^(-2)`
Using the relation, `v^(2) = u^(2) + 2as`,
`v^(2) = 2as = 2 xx 8 xx 10^(12) xx 2 xx 10^(-2) = 32 xx 10^(10) (because u = 0)`
`v = 4sqrt(2) xx 10^(5) ms^(-1)`