InterviewSolution
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InterviewSolution
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A fusion reaction of the type given below `._(1)^(2)D+._(1)^(2)D rarr ._(1)^(3)T+._(1)^(1)p+DeltaE` is most promissing for the production of power. Here D and T stand for deuterium and tritium, respectively. Calculate the mass of deuterium required per day for a power output of `10^(9) W`. Assume the efficiency of the process to be `50%`. Given : `" "m(._(1)^(2)D)=2.01458 am u," "m(._(1)^(3)T)=3.01605 am u` `m(._(1)^(1) p)=1.00728 am u` and `1 am u=930 MeV`. |
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Answer» Correct Answer - (a) `4MeV, 17.6 MeV` (b) `7.2 MeV` (c) `0.384 %` `2(._(1)^(2)D) rarr ._(1)^(3)T+._(1)^(1)P` Mass defect `DeltaM=M_("Product")-M_("Reactant")` `={(3.016049)+1.00785}-2[2.014102]` `=4.023899-4.028204` `implies Deltam=4.3xx10^(-3)` amu and `1` amu `rarr 931.5` `E=Deltamc^(2)=4.01 MeV` `.^(3)T_(1)+._(1)^(2)D rarr ._(2)^(4)He+._(0)^(1)n` `Deltam`(mass defect)`=DeltaM_("Product")-DeltaM_("Reactant")` `=[4.002603+1.008665]-[3.016049+2.014102]` `=[5.011268-5.030151]` `Deltam=0.018883` `E=Deltam(931.5) implies 17.58 MeV` `E_("deutron")=(DeltaE_("total"))/3=7.2 MeV` `("Total Energy")/("Total Mass")implies n=(Deltam_(1)+Deltam_(2))/(3_(1)^(2) D)` `implies n=((0.004305+0.018883)/(3(2.014102)))xx100=n=0.384%` |
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