A proton is fired with a velocity `7.45 xx 10^(5) m//s` towards another free proton at rest. Calculate the minimum distance of approach between the protons. The mass of a proton = `1.66 xx 10^(-22) kg`.

Explanation

Let us consider the moving proton to be the first proton and the free proton at rest to be the second proton.
The initial velocity of the first proton, `v = 7.45 xx 10^(5)m//s`. As the protons repel each other, the second proton starts moving away from the first proton. The velocity of the first proton gradually increases. When the velocities of the two protons become equal, the distance between them will be minimum. Let this velocity = v'
According to law of conservation of momentum
`mv+0 = mv' + mv'`
or, `v' = v/z`
Let the minimum distance of approach between the protons be r.
The electric potential energy of the protons when they are at a distance r apart = `1/(4 pi epsilon_0) cdot e/r`
According to the law of conservation of energy,
`1/2 mv^(2) + 0 = 1/(4 pi epsilon_0) cdot (epsilon^2)/r + 1/2 mv^('2) + 1/2 mv^('2)`
`1/(4 pi epsilon_0) cdot (e^2)/r + m(v/2)^(2)`
or, `1/(4 pi epsilon_0) cdot (e^2)/r = (mv^2)/(4)`
or, `r = 4/(mv^2) xx (e^2)/(4pi epsilon_0) = (4 xx (1.6 xx 10^(-19))^(2)xx9xx10^(9))/(1.66xx10^(-27)xx (7.45xx10^(5))^(2))`
`=1.0 xx 10^(-12)m`.