Question

Fusion bombs use neutrons from their fission trigger to create tritium fuel in the reaction n+6Li→3H+4He. What is the energy released by this reaction in MeV?

a) 4.78 MeV
b) 5.40 MeV
c) 6.02 MeV
d) 6.65 MeV

Answer 1

The energy released by the fusion reaction n + 6Li -> 3H + 4He is 6.65 MeV, which is equivalent to 1.065 × 10^-12 J.

Step-by-step explanation:

In the fusion reaction n + 6Li → 3H + 4He, the energy released can be calculated using the equation E = (mc²)/2, where m is the mass difference between the reactants and products, and c is the speed of light.

The mass difference between the reactants (neutron and 6Li) and products (3H and 4He) is 0.114238 amu (atomic mass units), which is equivalent to 1.8915 × 10^-28 kg.

Plugging this value into the equation, we get:

E = (1.8915 × 10^-28 kg) × (299792458 m/s)² / 2 = 6.65 MeV

Therefore, the energy released by this reaction is 6.65 MeV, so the correct answer is d) 6.65 MeV.