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

Let the kinetic energy of each deutron be K. Radius of deuteron = 2.0 fin =`2 xx 10^(-15)` m .
In a head-on collision, distance between the centres of two deuterons, `r=2 xx 2 xx 10^(-15)m = 4 xx 10^(-15)` m .
Let the initial energy of two deuterons be 2 K. This energy is converted into potential energy, on the head-on collision, of the deuterons and th at is given by:
`U=1/(4piepsilon_(0)).e^(2)/r`
From the law of conservation of energy, we have
`2K = 1/(4piepsilon_(0)).e^(2)/r`
or `K=1/(4piepsilon_(0)) xx e^(2)/(2r) =(9 xx 10^(9) xx (1.6 xx 10^(-19))^(2))/(2 xx 4 xx 10^(-16))`
`=2.88 xx 10^(-14)` J
`=1.80 xx 10^(5) eV = 180 keV`
`U =2 K = 180 xx 2 = 360 keV`.