Answer to Question #299358 in General Chemistry for Madey

Question

A proton and an electron are initially at rest at a distance of 9×10-¹⁰m. What will be their initial acceleration due to the electric force that they exert on each other?

Answer 1

$F=\frac{kq_1q_2}{r^2}$ $=\frac{9×10^9(1.6×10^{-19})^2}{(9×10^{-10})^2}$

$=2.844×10^{-10}$

$F=ma$

Acceleration of electron

$a=\frac{2.844×10^{-10}}{9.11×10^{-31}}$

$=3.117×10^{20}m/s^2$

Acceleration of proton

$a=\frac{2.844×10^{-10}}{1.67×10^{-27}}$

$=1.703×10^{17}m/s^2$