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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 eachother. |
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Answer» for head on collision, distance between centres of two deuterons=`r=2xxradius` `r=4fm=4xx10^(-15)m` charge of each deuteron `e=1.6xx10^(-19)C` Potential energy `=(e^2)/(4pi in_0r)=(9xx10^9(1.6xx10^(-19))^(2))/(4xx10^(-15))"joule"=(9xx1.6xx1.6xx10^(-14))/(4xx1.6xx10^(-16))KeV=360keV` As `P.E. =2xxK.E.` of each of deutron `=360keV` Coulomb barrier. `:.` K.E. of each deutron=`360/2=180keV` Thus is a measure of height of Coulomb barrier. |
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