Consider the nuclear reaction ${}^{3}_{1}$ H + ${}^{3}_{1}$ H → ${}^{4}_{2}$ He + 2 ${}^{1}_{0}$ n.The known atomic masses are ${}^{4}_{2}$ He: 4.002603 u ${}^{3}_{1}$ H: 3.016049u ${}^{1}_{0}$ n: 1.008665 u What is the energy released (positive)or absorbed (negative)in this nuclear reaction? (1 u = 931.5 MeV/c²)

Question

Quizplus · Accepted Answer

The total mass of the reactants (two ${}^{3}_{1}$H atoms) is 2(3.016049 u) = 6.032098 u.

The total mass of the products (${}^{4}_{2}$He + 2 ${}^{1}_{0}$n) is 4.002603 u + 2(1.008665 u) = 6.027933 u.

The difference in mass is:

6.032098 u - 6.027933 u = 0.004165 u

Using the conversion factor of 1 u = 931.5 MeV/c²,

the energy released is:

0.004165 u × 931.5 MeV/c²/u = 3.8851955 MeV

Rounding to the nearest tenth, the final answer is:

Energy released = 11.3 MeV (since the question asks for the absolute value of the energy, we do not include a negative sign)

Consider the Nuclear Reaction 13{}^{3}_{1}13​ H 13{}^{3}_{1}13​ H → 24{}^{4}_{2}24​ He + 2

Consider the Nuclear Reaction ${}^{3}_{1}$ H ${}^{3}_{1}$ H → ${}^{4}_{2}$ He + 2