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`.

Answer

(a) `4MeV, 17.6 MeV`
(b) `7.2 MeV` (c) `0.384 %`

Explanation

`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%`