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Calculate the height of the potential barrier for a head on collision of two deuterons. (Hint : The height of the potential barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other. Assume that they can be taken as hard spheres of radius 2.0 fm.) |
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Answer» Solution :As charge on each DEUTERON `q = 1.6 xx 10^(-19)` C and SEPARATION between their centres`= 2r=2xx2fm=4XX10^(-15)`m, hence their potential energy is : `U=1/(4piepsilon_(0)).((q)(q))/(2r)=1/(4piepsilon_(0)).q^(2)/(2r)=(9xx10^(9)xx(1.6xx10^(-19))^(2)/(4xx10^(-15))eV)=360keV`. This is the height of the potential barrier between the two deuterons. |
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