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Consider the so called D-T reaction (deu...

Consider the so called D-T reaction (deuterium-tritium fusion) `._1H^2+._1H^3to._2He^4+n`
Calculate the energy released in MeV in this reaction from the data
`m(._1H^2)=2.014102u, m(._1H^3)=3.016049u`
(b) Consider the radius of both deuterium and tritium to be approximately 2.0fm. what is the kinetic energy needed to overcome the Coulomb repulsion between the two nuclei? To what temperature must the gases the be heated to initiate the reaction?

Text Solution

Verified by Experts

(a) The given equation is
`""_(1)H^(2) + ""_(1)H^(3) to ""_(2)He^(4) +""_(0)n^(1) +Q`
So energy released in this reaction is given by
`Q=[m_(N)(""_(1)H^(2)) + m_(N)(""_(1)H^(3))-m_(N)(""_(2)He^(4)) -m_(n)]c^(2)`
`=[{m(""_(1)H^(2))+m(""_(1)H^(3))}=m(""_(2)He^(4))-m_(n)]c^(2)`
`=(2.014102 + 3.016049 - 4.002603 - 1.008665) u xx c^(2)`
=17.59 MeV
(b) Repulsive potential energy of two approaching nuclei when they are nearest to each other `(r_(0))` is:
`P.E. =1/(4piepsilon_(0)) xx e^(2)/r_(0)`
where, `r_(0) = 2r=2 xx 2.0 x 10^(-15)`m
`=4 xx 10^(-15)`m
So. P.E. `=(9 xx 10^(9)(1.6 xx 10^(-19))^(2))/(4.0 xx 10^(-15)) J = 5.76 xx 10^(-14)` J
Minimum kinetic energy required to overcome electrostatic repulsion is equal to the above­ calculated potential energy.
`rArr K.E. = 5.76 xx 10^(-14)` J
Since, `K.E. = 2(3/2k T)`
`therefore T =(K.E.)/(3k) = (5.76 xx 10^(-14))/(3 xx 1.38 xx 10^(-23)) = 1.391 xx 10^(9)` K.
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