Answer:
(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