Under the action of a given coulombic force the acceleration of an electron is `2.5 xx 10^(22) ms^(-1)`. Then, the magnitude of the acceleration of a proton under the action of same force is nearly

Options

A. `1.6xx10^(-19)ms^(-2)`
B. `9.1xx10^(31)ms^(-2)`
C. `1.5xx10^(19)ms^(-2)`
D. `1.6xx10^(27)ms^(-2)`

Answer

C

Explanation

The acceleration due to given coulombic force F is
`a=F/m oraprop1/m" ".... (i)`
`therefore" "a_(p)/a_(e)=m_(e)/m_(p)`, where `m_(e) and m_(p)` are masses of electron and proton respectively
`a_(p)=(a_(e)m_(e))/m_(p)=((2.5xx10^(22)ms^(-2))(9.1xx10^(-31)kg))/((1.67xx10^(-27kg)))`
`=13.6xx10^(18)ms^(-2)approx1.5xx10^(19)ms^(-2)`