Calculate the height of potential barrier for a head on collision of two deuterons. The effective radius of deuteron can be taken to be 2fm. Note that height of potential barrier is given by the Coulomb repulsion between two deuterons when they just touch each other.

Explanation

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 .