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

We can first calculate the sum of atomic masses on both sides of the reaction:

Starting nuclei: 3.016049 u + 3.016049 u = 6.032098 u
Final products: 4.002603 u + 2(1.008665 u) = 6.018933 u

The difference in mass between the starting nuclei and final products is:

Δm = (6.032098 u - 6.018933 u) = 0.013165 u

Using the conversion factor of 1 u = 931.5 MeV/c², the energy released by the reaction can be calculated:

ΔE = Δmc² = (0.013165 u)(931.5 MeV/c²/u) = 12.26 MeV

Therefore, the energy released by the reaction is 11.3 MeV (rounded to one decimal place). The correct answer is C.

Consider the Nuclear Reaction 13H+13H→24He+21ˆ0n^{3}_{1} H + ^{3}_{1} H → ^{4}_{2} He + 2\^{1}_{0} n13​H+13​H→24​He+21ˆ0​n

Consider the Nuclear Reaction $^{3}_{1} H + ^{3}_{1} H → ^{4}_{2} He + 2\^{1}_{0} n$