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. `1.6 xx 10^(-19)ms^(-2)`
B. `9.1 xx 10^(31)ms^(-2)`
C. `1.5 xx 10^(19)ms^(-2)`
D. `1.6 xx 10^(27)ms^(-2)`

Answer

C

Explanation

The acceleration of the electron due to given coulombic force F is `a_(e) = (F)/(m_(e))` …(i)
where `m_(e)` is the mass of the electron.
The acceleration of the proton due to same force F is
`a_(p) = (F)/(m_(p))` …(ii)
where `m_(p)` is the mass of the proton.
Divide (ii) by (i), we get `(a_(p))/(a_(e)) = (m_(e))/(m_(p))`
`a_(p) = (a_(e)m_(e))/(m_(p)) = ((2.5 xx 10^(22) ms^(-2))(9.1 xx 10^(-31)kg))/((1.67 xx 10^(-27) kg))`
`~~ 1.5 xx 10^(19) ms^(-2)`