Answer
The acceleration of the proton is $3.4 \times 10^{18} m/s^2$ and that of the electron is $6.3 \times 10^{21} m/s^2$.
Work Step by Step
The proton and electron have equal and opposite charges, so they will attract each other. The magnitude of the force of attraction is given by Coulomb's law –
$ F = \frac{1}{4 \pi \epsilon_{0}} \frac{|q_1 q_2|}{r^2} = \frac{1}{4 \pi \epsilon_{0}} \frac{e^2}{r^2}$
. . . where $e = 1.6 \times 10^{-19} C$. Substituting values, we get –
$ F = \frac{(9\times 10^9)(1.6\times10^{-19})^2}{(2.0 \times 10^{-10})^{2}} = 5.76 \times 10^{-9}N$
From Newton's second law, $F = ma$. Therefore, for each particle, $a = F/m$.
$a_{proton} = \frac{F}{m_{proton}} = \frac{5.76 \times 10^{-9}}{1.67 \times 10^{-27}} = 3.5 \times 10^{18} m/s^2$
$a_{electron} = \frac{F}{m_{electron}} = \frac{5.76 \times 10^{-9}}{9.11 \times 10^{-31}} = 6.3 \times 10^{21} m/s^2$
