Question

If a proton and an electron are released when they are 5.50×10−10 mm apart (typical atomic distances), find the initial acceleration of each of them. Express your answer in meters per second squared.

Answer 1

Answer:

$a_e=8.376\times 10^{26}\,ms^{-2}\\\\a_p=4.56\times 10^{23}\,ms^{-2}$

Explanation:

Distance between proton and electron = r = 5.5 x 10⁻¹³ m

Magnitude of charge on electron and proton = 1.67 x 10⁻¹⁹ C

To find their initial acceleration first force between them is:

$F= \frac{1}{4\pi \epsilo_{o}}\frac{q_pq_e}{r^2}\\\\F= (9\times 10^9)\frac{(1.60\times 10^{-19})^2}{(5.5\times 10^{-13})^2}\\\\F=7.63\times 10^{-4}N$

Initial acceleration of electron and proton are found using

F=ma

a=F/m

$a_e=\frac{F}{m_e}\\\\a_e=\frac{7.63\times 10^{-4}}{9.109 \times 10^{-31}}\\\\a_e=8.376\times 10^{26}\,ms^{-2}\\ \\\\a_{p}=\frac{7.63\times 10^{-4}}{1.673\times 10^{-27}}\\\\a_p=4.56\times 10^{23} \,ms^{-2}$