Consider the following fusion reaction .The mass of is 3.016029 u,the mass of is 4.002602 u,and the mass of is 7.0169298 u.Assuming that the Be atom is at rest after the reaction,calculate the minimum possible energy of the photon g that is released in this reaction.

Question

Consider the following fusion reaction  .The mass of  is 3.016029 u,the mass of  is 4.002602 u,and the mass of  is 7.0169298 u.Assuming that the Be atom is at rest after the reaction,calculate the minimum possible energy of the photon g that is released in this reaction.

Quizplus · Accepted Answer

The energy released in the reaction is equal to the difference in mass between the reactants and products, according to Einstein's famous equation E= mc^2.
 
Δm = (3.016029 u + 4.002602 u) - 7.0169298 u
Δm = 0.0017014 u

The energy released is given by:
E = Δm * c^2

where c is the speed of light.

E = 0.0017014 u * (1.66054 x 10^-27 kg/u) * (2.998 x 10^8 m/s)^2
E = 1.602 x 10^-13 J

We can convert this to MeV using the conversion factor 1.602 x 10^-13 J/MeV:

E = 1.602 x 10^-13 J / (1.602 x 10^-13 J/MeV)
E = 1 MeV (to one significant figure)

Therefore, the minimum possible energy of the photon released in this reaction is 1 MeV, which corresponds to choice (C).

Consider the Following Fusion Reaction