Answer:
The distance of the closest approach is 78.455 fm
Solution:
As per the question:
Charge on [tex]\alpha-particle[/tex], q = + 2e
Charge on the nucleus of gold atom, q' = + 79e
Kinetic Energy, KE = 2.9 MeV = [tex]2.9\times 10^{6} eV[/tex]
e = [tex]1.6\times 10^{-19}\ C[/tex]
Now,
We know that the electrostatic potential energy,
PE = [tex]\frac{1}{4\pi\ e[psilon_{o}.\frac{qq'}{r}}[/tex]
To calculate the distance of the closest approach, r:
PE = KE
[tex]9\times 10^{9}\times \frac{2e\times 79e}{r}} = 2.9\times 10^{6}e[/tex]
[tex]r = \frac{9\times 2\times 79\times 1.6\times 10^{- 19}}{2.9\times 10^{6}}[/tex]
[tex]r = 7.8445\times 10^{- 14}\ m = 78.455\ fm[/tex]
