What energy must be added or given off in a reaction where two hydrogen atoms and two neutrons are combined to form a helium atom? (Atomic masses for each: hydrogen,1.007 825 u;neutron,1.008 665 u;helium,4.002 602 u;also,1 u = 931.5 MeV/c²)

Question

Quizplus · Accepted Answer

In the reaction described, two hydrogen atoms and two neutrons combine to form a helium atom. The mass of two hydrogen atoms and two neutrons is (2*1.007825u)+(2*1.008665u) = 4.03398u. The mass of a helium atom is 4.002602u. The difference in mass between the reactants and the product is 0.031378u. Using the conversion factor given, 1u = 931.5 MeV/c², this difference in mass can be converted to energy:
0.031378u × 931.5 MeV/c² = 29.29 MeV
However, we must also take into account that the reaction involves the combining of particles, and so we must subtract the rest energy of the particles from this result. This can be found by multiplying the number of particles by their respective rest energies:
(2+2) × 0.511 MeV = 2.044 MeV
Therefore, the net energy released in the reaction is:
29.29 MeV - 2.044 MeV = 27.246 MeV
This is closest to the answer choice C, which represents 28.3 MeV given off (we must round to the nearest tenth of an MeV).

What Energy Must Be Added or Given Off in a Reaction