when two deuterons collide head - on , the distance between their centres d is given as :
Radius of `1 ^(st)` deuteron + Radous of `2^(nd)` deuteron
radius of a deuertiopn nucleus `=2 fm =2xx10^(-15)m`
` therefore d=2xx10^(-15)+2xx10^(-15)=4xx10^(-15)`m
change on a deuteron nuclear = change on an electron `=e=1.6 10^(-19)`C
potential energy of the two - deuteron syatem :
`V=(e^(2))/(4 piin_(@)d)`
where
`in_(@)=` permitticvity of free space
`(1)/(4piin_(@))=9xx10^(9)Nm^(2)C^(-2)`
`therefore v =(9xx10^(9)xx(1.6xx10^(-19))^(2))/(4xx10^(-15))J`
`=(9xx10^(9) xx(1.6 xx10^(-19))^(2))/(4xx10^(-15))xx(1.6xx10^(-19))EV`
=360 keV
Hence the height of the potential barrier of the two - deuteroun system is 360 keV .