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Following fusion reaction is to be used ...

Following fusion reaction is to be used for the production of power.
`""_(1)H^(2) + ""_(1)H^(2) to ""_(1)H^(3) + ""_(1)H^(1)`
Calculate requirem ent of mass of deuterium per day if power is produced at a rate of `10^9` W. Assume that the above process is used w ith efficiency of 50%.
`m(""_(1)H^(2)) = 2.01458` amu
`m(""_(1)H^(3)) = 3.01605` amu
`m(""_(1)H^(1)) = 1.00728` amu
1 amu `=931 MeV//c^(2)`.

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To solve the problem step by step, we will follow these steps: ### Step 1: Calculate the mass defect (Δm) The mass defect is the difference between the mass of the reactants and the mass of the products in the fusion reaction. **Reactants:** - 2 Deuterium nuclei: \(2 \times m(^{2}H) = 2 \times 2.01458 \, \text{amu} = 4.02916 \, \text{amu}\) ...
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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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