Two protons are relased when they are initially 2 fermi apart. Find their seepds when they are 4 fermi apart. Give the mass of a proton. `=1.67 xx 10^(-27)kg` and charge `=1.6 xx 10^(-19)C`.

Explanation

Data supplied,
From the law of conservation of energy,
Gain in KE =Loss in P.E.
Since the protons are intially at rest, gain in KE `=1/2 mv^(2)=1/2 (2m_(p))v^(2)=m_(p)v^(2)`
`m_(p)=1.67 xx 10^(-27)kg`
Loss in PE `=(1)/((4pi epsi_(0)) q_(1)q_(2)) ((r_(2)-r_(1)))/(r_(1)r_(2)), r_(1)="2 fermi" =2 xx 10^(-15)m`
`r_(2)="4 fermi" =4 xx 10^(-15)m, q_(1)=q_(2)=e=1.6 xx 10^(-19)C`
`m_(p)v^(2)=(1)/(4pi epsi_(0)) e^(2) (r_(2)-r_(1))/(r_(1)r_(2))`
`v^(2) =(1)/(4pi epsi_(0)) e^(2)/m_(p) ((r_(2)-r_(1)))/(r_(1)r_(2))=9 xx 10^(9) xx ((1.6 xx 10^(-19))^(2))/(1.67 xx 10^(-27)) xx ((4-2) xx 10^(-15))/(2 xx 10^(-15) xx 4 xx 10^(-15))=34.49 xx 10^(12)`
Speed `v=5.87 xx 10^(6) m//s`