What energy must be added or given off in a reaction where a helium atom is separated into two hydrogen atoms and two neutrons? (Atomic masses for each: hydrogen,1.007825 u;neutron,1.008665 u;helium,4.002602 u;also,1 u = 931.5 MeV/c². ) A)28.3 MeV given off B)20.7 MeV added C)20.7 MeV given off D)28.3 MeV added

Question

Quizplus · Accepted Answer

To find the energy involved in separating a helium atom into two hydrogen atoms and two neutrons, we calculate the mass defect and then convert it to energy. The mass of the helium atom is 4.002602 u. The combined mass of the products (two hydrogen atoms and two neutrons) is (2 * 1.007825 u) + (2 * 1.008665 u) = 4.03322 u. The mass defect is the difference between the initial and final masses, which is 4.03322 u - 4.002602 u = 0.030618 u. Converting this mass defect into energy using E=mc^2 (where 1 u = 931.5 MeV/c^2), we get 0.030618 u * 931.5 MeV/u = 28.53 MeV. Since the final mass is greater than the initial mass, energy must be added to the system to make this transformation happen.

What Energy Must Be Added or Given Off in a Reaction