An `alpha` 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 ennd of `2xx10^(-2)m` path when it starts from rest is

Options

A. `2sqrt(3)xx10^(5)ms^(-1)`
B. `8xx10^(5)ms^(-1)`
C. `16xx10^(5)ms^(-1)`
D. `4sqrt(2)xx10^(5)ms^(-1)`

Answer

D

Explanation

Here mass of `alpha` -particle `m_(alpha)=6.4xx10^(-27)kg`
Charge of `alpha`- particle `q_(alpha)=3.2xx10^(-19)C`
Electric field `E=1.6xx10^(5)Vm^(-1)`
Force on `alpha`- particle in a uniform electric field E is
`F=q_(alpha)E=3.2xx10^(-19)xx1.6xx10^(5)N`
`=3.2xx1.6xx10^(-14)N`
Acceleration of the `alpha`- particle is
`a=F/(m_(alpha))=(3.2xx1.6xx10^(-14))/(6.4xx10^(-27))ms^(-2)`
Given `u=0 `
`:.v^(2)=2aS=(2xx3.2xx1.6xx10^(-14)xx2xx10^(-2))/(6.4xx10^(-27))`
(As `S=2xx10^(-2)m`)
`=3.2xx10^(11)=32xx0^(10)`
or `v=4sqrt(2)xx10^(5)ms^(-1)`
Alternate soltion :
According to work energy theorem,
`1/2m_(alpha)v^(2)-1/2m_(alpha)(0)^(2)=q_(alpha)ES`
or `1/2m_(alpha)v^(2)=q_(alpha)ES` or `v=sqrt((2q_(alpha)ES)/(m_(alpha)))`
`v=sqrt((2xx3.2xx10^(-19)xx1.6xx10^(5)xx2xx10^(-2))/(6.4xx10^(-27))`
`v=sqrt(3.2xx10^(11))=sqrt(32xx10^(10))=4sqrt(2)xx10^(5)m//s`