Question

An alpha particle (the nucleus of a helium atom) is at rest at the origin of a Cartesian coordinate system. A proton is moving with a velocity of v towards the alpha particle in the xˆ direction. If the proton is initially far enough away to have no potential energy, how close does the proton get to the alpha particle? Your answer should be in terms of v, mp (mass of a proton) and e (charge of a proton).

Answer 1

The distance of closest approach between proton and an alpha particle is
`(e^(2))/(M\pi\epsilon_(o)v_(p)^(2))`

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
`v_(p)` and mass M with the charge on proton as 'e' and that on
`\alpha`- particle as 2e.

Therefore,

Kinetic Energy of proton, KE must be equal to the its Electrostatic Potential Energy and is given as:

`(1)/(2)Mv_(p)^(2) = (1)/(4\pi\epsilon_(o))(e* 2e)/(R)}`

where

e = electronic charge

R = separation distance between the two charges

`R = (e^(2))/(M\pi\epsilon_(o)v_(p)^(2))`