Question

3. In the following deuterium reaction, the reaction energy is as stated: ¹N(d,p) ¹N Q-8.61 MeV ¹5N(d, a) ¹3C Q-7.68 MeV 13C (d, a)¹B Q-5.16 MeV ¹B (a, n) ¹4C Q=? Given that He = 4.002603u, H = 2.014102u, H=1.007825u and n=1.008665u, (4marks) calculate the Q value of the four reaction.​

Answer 1

The energy released in joules for the fusion reaction 2H + 2H -> 3He + n is approximately 7.20 x 10^-14 J.

Step-by-step explanation:

To find the energy released in joules for the fusion reaction 2H + 2H -> 3He + n, we need to calculate the difference in mass between the reactants and the products. The mass of 2H is 2.014102u and the mass of 3He is 3.016029u. The mass of n (neutron) is 1.008665u. The difference in mass is (3.016029u + 1.008665u) - (2.014102u + 2.014102u) = 0.00249u.

To convert this mass difference to energy, we use the equation E = mc^2, where m is the mass difference and c is the speed of light (approximately 3 x 10^8 m/s). So, E = (0.00249u)(1.6605 x 10^-27 kg/u)(3 x 10^8 m/s)^2 = 7.20 x 10^-14 J.