Question

If a proton and an electron are released when they are 5.50×10−10 m apart (typical atomic distances), find the initial acceleration of each of them.

Answer 1

We know the initial force of atrction F = kq1q2 / r^2

where k is the Coulomb constant, and q1 and q2 are actually the same charge, that of the electron and proton e = 1.6x10^-19 C and r is the distance separating the particles.

We also know that F = ma

so we can equate the two terms to get

Accelerattion of electron

m_ea_e = ke^2 / r^2
or
a_e = ke^2 / m_er^2 = [9 x 10⁹ x (1.6 x 10^-19)²] / [9.1 x 10^-31 x (5.5 x 10^-10)²]

a_e = 8.4 x 10²⁰ m/s²

Accelerattion of proton

m_pa_p = ke^2 / r^2
or
a_p = ke^2 / m_pr^2 = [9 x 10⁹ x (1.6 x 10^-19)²] / [1.67 x 10^-27 x (5.5 x 10^-10)²]

a_p = 4.6 x 10¹⁷ m/s²