The rest mass of a proton is `1.67xx10^(-27)` kg. the kinetic energy of proton when it moves with a speed of `2.1xx10^(8)` m/s is

Options

A. `6xx10^(-12) J`
B. `5xx10^(-11) J`
C. `6xx10^(-11)J`
D. `7xx10^(-11) J`

Answer

C

Explanation

`m=m_(0)sqrt(1-v^(2)/c^(2))=m_(0)/sqrt(1-((2.1xx10^(8))/(3xx10^(8)))^(2))=m_(0)/sqrt(0.51)`
`m=(1.67xx10^(-27))/0.7141=2.338xx10^(-27)` kg
Now `E_(k)=E-E_(0)`
`=mc^(2)-m_(0)c^(2)`
`=(m-m_(0))c^(2)`
`=(2.338-1.670)xx10^(-27)xx9xx10^(16)`
`=0.668xx9xx10^(-11)`
`=6.01xx10^(-11) J`