Under the action of a given coulombic force, the acceleration of an electron is 2.5 x 10²² ms^-2. Then, the magnitude of the acceleration of a proton under the action of same force is nearly

The acceleration of the electron due to given coulombic force, F is  $a_e=\frac{F}{m_e}$   …(i)

where, m_e is the mass of the electron.
The acceleration of the proton due to same force F is

$a_p=\frac{F}{m_p}$         …(ii)

where, m_p is the mass of the proton

On dividing Eq. (ti) by Eq. (i), we get  $\frac{a_p}{a_e}=\frac{m_e}{m_p}$

$a_p=\frac{a_e{m}_e}{m_p}=\frac{\left(2.5\times {10}^{22}{ \mathrm{ms}}^{-2}\right)\left(9.1\times {10}^{-31} \mathrm{kg}\right)}{\left(1.67\times {10}^{-27} \mathrm{kg}\right)}$

$=13.6\times {10}^{18}{ \mathrm{ms}}^{-2}\approx 1.5\times {10}^{19}{ \mathrm{ms}}^{-2}$