The nuclear reaction `._(1)^(2)H + ._(1)^(2)H to ._(2)^(4)He` is called

Calculate the energy released in joules in the following nuclear reaction : ""_(1)^(2)H + ""_(1)^(2)H to ""_(2)^(3)He + ""_(0)^(1)N Assume that the masses of ""_(1)^(2)H, ""_(2)^(3)He and neutron are 2.0141, 3.0160 and 1.0087 amu respectively.

._(1)^(3)H and ._(2)^(4)He are:

Calculate the energy in the reaction 2 ._(1)^(1)H + 2 ._(0)^(1)n to ._(2)^(4)He Given, H = 1.00813 amu, n = 1.00897 amu and He = 4.00388

In a typical nuclear reaction, e.g. ""_(1)^(2) H+ ""_(1)^(2) H to ""_(2)^(3) He +n + 3.27 Mev, although number of nucleons is conserved, yet energy is released. How ? Explain.

Calculate the energy released in the following nuclear reaction: ""_(1)^(2)H+ ""_(1)^(2)H =""_(2)^(4)He Mass of ""_(1)^(2)H = 2.01419u, Mass of ""_(2)^(4)He =4.00277u

Calculate the energy released in joules and MeV in the following nuclear reaction: ._(1)H^(2) + ._(1)H^(2) rarr ._(2)He^(3) + ._(0)n^(1) Assume that the masses of ._(1)H^(2) , ._(2)He^(3) , and neutron (n) , respectively, are 2.40, 3.0160, and 1.0087 in amu.

Calculate the energy released in joules and MeV in the following nuclear reaction: ._(1)H^(2) + ._(1)H^(2) rarr ._(2)He^(3) + ._(0)n^(1) Assume that the masses of ._(1)H^(2) , ._(2)He^(3) , and neutron (n) , respectively, are 2.40, 3.0160, and 1.0087 in amu.