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For the `D-T` fusion reaction, find the rate at which deuterium & trithium are consumed to produce `1 MW`. The `Q`-value of `D-T` reactions is `17.6 MeV` & assume all the energy from the fusion rection is available. |
Answer» Correct Answer - `(2)/(N_(A))xx(1)/(17.6e)xx10^(-3) Kg//s= 1.179xx10^(-9)kg//s` `(3)/(N_(A))xx(1)/(17.6e)xx10^(-3)kg//s = 1.769xx10^(-9) kg//s` Each deiterium nucleus prodeuces `17.6 MeV`. `1 kg` of deuterium `=(1xx10^(3))/(2)N_(A)` no. of deuterium `-=17.6xx(10^(3))/(2)N_(A) MeV` energy produced. `:.` To produce `1 MW`, amount of deuterium in `kg` required per second. `=(1xx10^(6))/(17.6xx(10^(3))/(2)xxN_(A)xxexx10^(6))kg//s=17.69xx10^(-9)kg//s` |
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