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

`a_e=8.376* 10^(26)\,ms^(-2)\\\\a_p=4.56* 10^(23)\,ms^(-2)`

Step-by-step 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= (1)/(4\pi \epsilo_(o))(q_pq_e)/(r^2)\\\\F= (9* 10^9)((1.60* 10^(-19))^2)/((5.5* 10^(-13))^2)\\\\F=7.63* 10^(-4)N`

Initial acceleration of electron and proton are found using

F=ma

a=F/m

`a_e=(F)/(m_e)\\\\a_e=(7.63* 10^(-4))/(9.109 * 10^(-31))\\\\a_e=8.376* 10^(26)\,ms^(-2)\\ \\\\a_(p)=(7.63* 10^(-4))/(1.673* 10^(-27))\\\\a_p=4.56* 10^(23) \,ms^(-2)`