Answer:
a=2.304×10¹⁶m/s²
Explanation:
Given data
Distance d=2.5 nm=2,5×10⁻⁹m
Mass of proton m=1.6×10⁻²⁷kg
charge of proton q=1.6×10⁻¹⁹C
To find
acceleration a
Solution
Apply the Coulombs Law
[tex]F=k\frac{q_{1}q_{2} }{r^{2} }[/tex]
Where k is coulombs constant (k=9×10⁹Nm²/C²)
q=q₁=q₂
r=d
So
[tex]F=k\frac{|q^{2} |}{d^{2} }\\ as \\F=ma\\ma=k\frac{|q^{2} |}{d^{2} }\\a=\frac{k}{m} \frac{|q^{2} |}{d^{2} }\\a=\frac{(9*10^{9} )*(1.6*10^{-19} )^{2} }{(1.6*10^{-27} )*(2.5*10^{-9} )^{2} }\\ a=2.304*10^{16}m/s^{2}[/tex]
