Question

If a proton and an electron are released when they are 7.00×10−10 m apart (typical atomic distances), find the initial acceleration of each of them. aelectron = nothing m/s2

Answer 1

Answer:

The acceleration of the proton is 2.823 x 10¹⁷ m/s²

The acceleration of the electron is 5.175 x 10²⁰ m/s²

Explanation:

Given;

distance between the electron and proton, r = 7 x 10⁻¹⁰ m

mass of proton, $m_p$ = 1.67 x 10⁻²⁷ kg

mass of electron, $m_e$ = 9.11 x 10⁻³¹ kg

The attractive force between the two charges is given by Coulomb's law;

$F = \frac{k(q_p)(q_e)}{r^2}$

where;

k is Coulomb's constant = 9 x 10⁹ Nm²/c²

$F = \frac{k(q_p)(q_e)}{r^2} \\\\F = \frac{(9*10^9)(1.602*10^{-19})(1.602*10^{-19})}{(7*10^{-10})^2} \\\\F = 4.714 *10^{-10} \ N$

Acceleration of proton is given by;

F = ma

$F = m_pa_p\\\\a_p = \frac{F}{m_p}\\\\a_p = \frac{4.714*10^{-10}}{1.67*10^{-27}}\\\\a_p = 2.823 *10^{17} \ m/s^2$

Acceleration of the electron is given by;

$F = m_ea_e\\\\a_e = \frac{F}{m_e}\\\\a_e = \frac{4.714*10^{-10}}{9.11*10^{-31}}\\\\a_e = 5.175 *10^{20} \ m/s^2$