Respuesta :
Answer:
The distance of closest approach between proton and an alpha particle is [tex]\frac{e^{2}}{M\pi\epsilon_{o}v_{p}^{2}}[/tex]
Solution:
As per the question:
In order to satisfy the given condition in the question, we apply the law of conservation in the given case.
Suppose the Velocity of the proton is [tex]v_{p}[/tex] and mass M with the charge on proton as 'e' and that on [tex]\alpha[/tex]- particle as 2e.
Therefore,
Kinetic Energy of proton, KE must be equal to the its Electrostatic Potential Energy and is given as:
[tex]\frac{1}{2}Mv_{p}^{2} = \frac{1}{4\pi\epsilon_{o}}\frac{e\times 2e}{R}}[/tex]
where
e = electronic charge
R = separation distance between the two charges
[tex]R = \frac{e^{2}}{M\pi\epsilon_{o}v_{p}^{2}}[/tex]
