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Extimate the amount of energy released i...

Extimate the amount of energy released in the following nuclear fusion reaction. `overset(2)(1)H+overset(2)(1)Hrightarrowoverset(3)(2)He+overset(1)(0)n` Given mass of `overset(2)(1)H=2.0141` amu, mass of `overset(3)(2)He=3.0160` amu, mass of `overset(1)(0)n=1.0087` amu and 1 amu =`1.661xx10^(-27)kg`. Express your answer in units of MeV?

Text Solution

Verified by Experts

The correct Answer is:
3.27
`""_(1)^(2) H + ""_(1)^(2)H rarr ""_(2)^(3)He + ""_(0)^(1)n`
Total initial mass `(""_(1)^(2)H + ""_(1)^(2)H)`
= 2.0141 + 2.0141= 4.0282 amu
Total final mass `(""_(2)^(3)He+ ""_(0)^(1)n)`
`=3.0160 + 1.0087= 4.0247` amu
Decrease in mass, `Delta m= 4.0282 - 4.0247 = 0.0035` amu
`=0.0035 xx 1.661 xx 10^(-27) kg`
`therefore` Energy released `= Delta m= c^(2) = 0.0035 xx 1.661 xx 10^(-27) xx (3 xx 10^(8))^(2)`
`=5.232 xx 10^(-13) J= (5.232 xx 10^(-13))/(1.6 xx 10^(-13))= 3.27MeV`
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