An electron is accelerated through a potential difference of 45.5 volt. The velocity acquired by it is (in `ms^(-1)`)

Options

A. `10^(6)`
B. zero
C. `4xx10^(6)`
D. `4xx10^(4)`

Answer

C

Explanation

`(1)/(2)mv^(2)=ev`
`rArr` `v^(2)=(45.5xx1.6xx10^(-19)xx2)/(9.1xx10^(-31))=16xx10^(12)`
`rArr` v=`4xx10^(6)m//s`.