Question

If a proton and an electron are released when they are 2.50×10^-10m apart (typical atomic distances), find the initial acceleration of each of them.
a)Acceleration of electron
b)Acceleration of proton

Answer 1

To solve this exercise, we will first proceed to calculate the electric force given by the charge between the proton and the electron (it). From the Force we will use Newton's second law that will allow us to find the acceleration of objects. The Coulomb force between two charges is given as

$F = k \frac{q_1q_2}{r^2}$

Here,

k = Coulomb's constant

q = Charge of proton and electron

r = Distance

Replacing we have that,

$F = (9*10^9)(\frac{(1.602*10^{-19})^2}{2.5*10^{-10}})$

$F = 3.6956*10^{-9}N$

The force between the electron and proton is calculated. From Newton's third law the force exerted by the electron on proton is same as the force exerted by the proton on electron.

The acceleration of the electron is given as

$a_e = \frac{F}{m_e}$

$a_e = \frac{3.6956*10^{-9}}{9.11*10^{-31}}$

$a_e = 4.0566*10^{21}m/s^2$

The acceleration of the proton is given as,

$a_p = \frac{F}{m_p}$

$a_p = \frac{3.6956*10^{-9}}{1.672*10^{-27}}$

$a_p = 2.21*10^{18}m/s^2$