Saved Bookmarks
| 1. |
Consider the D - T reaction (deuterium - tritium fusion) ._(1)^(2)H+ ._(1)^(3)H to ._(2)^(4)He + n Consider the radius of both deuterium and tritium to be approximately 2.0 fm. What is the kinetic energy needed to overcone the Coluomn repulsion between the two nuclei ? To what temperature must the gas heated to initiate the reaction ? |
|
Answer» Solution :HINT : KINETIC energy required for one fusion event = average thermal kinetic energy available with the interacting particles = 2(3kt/2) , k = Boltzman's CONSTANT, T = absolute temperature. Repulsive potential energy of two nuclei when they almost touch each other is `= (q^(2))/(4pi epsilon_(0)(2r))=(9xx10^(9)(1.6xx10^(-19))^(7))/(2xx2xx10^(-15))` Joule `= 5.76xx10^(-14)` Joule Classically KE atleast equal to this amount is required to overcome Coulomb repulsion. Using the relation `K.E. = 2xx(3)/(2)KT` `T=(K.E)/(3k)=(5.76xx10^(_14))/(3xx1.38xx10^(-23))=1.39xx10^(9)K` In actual PRACTISE the temperature required for trigerring the reaction is somewhat less. |
|