If a proton and an electron are released when they are 6.50×10⁻¹⁰ m apart (typical atomic distances), find the initial acceleration of each of them.
a) acceleration of electron
b) acceleration of proton

(a) Acceleration of electron= 5.993×10²⁰ m/s²

(b) Acceleration of proton= 3.264×10¹⁷ m/s²

Explanation:

Given Data

distance r= 6.50×10⁻¹⁰ m

Mass of electron Me=9.109×10⁻³¹ kg

Mass of proton Mp=1.673×10⁻²⁷ kg

Charge of electron qe= -e = -1.602×10⁻¹⁹C

Charge of electron qp= e = 1.602×10⁻¹⁹C

To find

(a) Acceleration of electron

(b) Acceleration of proton

Solution

Since the charges are opposite the Coulomb Force is attractive

So

From Newtons Second Law of motion

F=ma

a=F/m

For (a) Acceleration of electron

For(b) Acceleration of proton

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