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 xx10^(-19) m//s^2`
B. `9.1 xx10^(31) m//s^2`
C. `1.5 xx10^(19) m//s^2`
D. `1.6 xx10^(-27) m//s^2`

Answer

C

Explanation

`a_e = (eE)/(m_e) = 2.5 xx10^(22) m//s`
`a_P = (eE)/(m_p) , a_e/(a_p) = (m_P)/(m_e) = (1.67 xx10^(-27)kg)/(9.1 xx10^(-31)kg)`
`implies a_p = (2.5 xx10^(22) xx 9.1 xx10^(-31))/(1.67 xx10^(-27))`
`=1.3 xx10^(19)m//s^2 ~~ 1.5 xx10^(19) m//s^2`