How much energy (in MeV) is released in the reaction proton + deuteron goes to 3He + gamma? (You may ignore the initial kinetic energy of the reactants.)
For the purposes of this problem, use the following masses (in MeV/c2):
Proton: 938.783263054
Neutron: 939.5657182828
Deuteron: 1876.12457290064
3H: 2809.43251234168
3He: 2809.41388245528
To find the energy released in the given nuclear reaction, we need to use the principle of mass-energy equivalence, which is represented by Einstein's famous equation E = mc^2. In this equation, E represents the energy, m represents the mass, and c represents the speed of light.
To apply this equation, we need to calculate the initial and final masses and find the difference.
Given masses (in MeV/c^2):
Proton (p) = 938.783263054
Deuteron (d) = 1876.12457290064
3He (helium-3) = 2809.41388245528
The mass of the proton and deuteron is the same before and after the reaction since they are individually conserved and not part of the reaction products.
Now, let's calculate the initial and final masses of the reaction products:
Initial mass = mass of the proton (p) + mass of the deuteron (d)
= 938.783263054 + 1876.12457290064
= 2814.90783595464 MeV/c^2
Final mass = mass of 3He (helium-3)
= 2809.41388245528 MeV/c^2
To find the energy released, we subtract the final mass from the initial mass and multiply it by the speed of light squared:
Energy released (E) = (Initial mass - Final mass) * c^2
where c is the speed of light, approximately 2.998 × 10^8 meters per second.
Energy released (E) = (2814.90783595464 - 2809.41388245528) * (2.998 × 10^8)^2
Calculating the above expression, we find:
Energy released (E) ≈ 978738.6 MeV
Therefore, approximately 978738.6 MeV of energy is released in the given reaction.