The mass of a hydrogen atom (\(_{\rm{1}}^{\rm{1}}{\rm{H}}\)) is\({\rm{1}}{\rm{.007825}}\)amu; that of a tritium atom (\(_{\rm{1}}^{\rm{3}}{\rm{H}}\)) is\({\rm{3}}{\rm{.01605}}\)amu; and that of an\({\rm{\alpha }}\)particle is\({\rm{4}}{\rm{.00150}}\)amu. How much energy in kilojoules per mole of\(_{\rm{4}}^{\rm{2}}{\rm{He}}\)produced is released by the following fusion reaction:\(_{\rm{1}}^{\rm{1}}{\rm{H + }}_{\rm{1}}^{\rm{3}}{\rm{H}} \to _{\rm{4}}^{\rm{2}}{\rm{He}}\).

Nuclear chemistry is a branch of chemistry that studies radioactivity, nuclear processes, and atomic nuclei alterations such as nuclear transmutation and nuclear characteristics.

We can get energy by following a simple formula:

\({\rm{E = m \times }}{{\rm{c}}^{\rm{2}}}\)

But first, we must understand the response that takes place:

\(_{\rm{1}}^{\rm{1}}{\rm{H + }}_{\rm{1}}^{\rm{3}}{\rm{H}} \to _{\rm{4}}^{\rm{2}}{\rm{He}}\)

The total mass then will be:

\(\begin{array}{c}{\rm{mass = m( reactants ) - m( products )}}\\{\rm{mass = (1}}{\rm{.007285amu + 3}}{\rm{.01605amu) - 4}}{\rm{.00150amu}}\\{\rm{mass = 0}}{\rm{.022375amu}}\end{array}\)

Then, converting the value of amu in kg is:

\(\begin{array}{c}{\rm{1amu = 1}}{\rm{.66 \times 1}}{{\rm{0}}^{{\rm{ - 27}}}}{\rm{\;kg}}\\{\rm{ mass = 0}}{\rm{.022375amu \times 1}}{\rm{.66 \times 1}}{{\rm{0}}^{{\rm{ - 27}}}}{\rm{\;kg \times am}}{{\rm{u}}^{{\rm{ - 1}}}}\\{\rm{ = 3}}{\rm{.715 \times 1}}{{\rm{0}}^{{\rm{ - 29}}}}{\rm{\;kg}}\end{array}\)

The energy is then evaluated as:

\(\begin{array}{c}{\rm{E = m \times }}{{\rm{c}}^{\rm{2}}}\\{\rm{E = 3}}{\rm{.715 \times 1}}{{\rm{0}}^{{\rm{ - 29}}}}{\rm{\;kg \times }}{\left( {{\rm{2}}{\rm{.99 \times 1}}{{\rm{0}}^{\rm{8}}}{\rm{\;m}}{{\rm{s}}^{\rm{ - }}}{\rm{1}}} \right)^{\rm{2}}}\\{\rm{E = 3}}{\rm{.322 \times 1}}{{\rm{0}}^{{\rm{ - 12}}}}{\rm{\;J}}\\{\rm{E = 3}}{\rm{.322 \times 1}}{{\rm{0}}^{{\rm{ - 15}}}}{\rm{\;kJ}}\end{array}\)

However, that is energy per nucleus. Multiply it by the Avogadro number to find the energy per mole:

\(\begin{array}{c}{\rm{E = 3}}{\rm{.322 \times 1}}{{\rm{0}}^{{\rm{ - 15}}}}{\rm{\;kJ \times nucleu}}{{\rm{s}}^{{\rm{ - 1}}}}{\rm{ \times 6}}{\rm{.022 \times 1}}{{\rm{0}}^{{\rm{23}}}}{\rm{nucleus \times mo}}{{\rm{l}}^{{\rm{ - 1}}}}\\{\rm{E = 2 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{kJmo}}{{\rm{l}}^{{\rm{ - 1}}}}\end{array}\)

Therefore, the energy is: \({\rm{E = 2 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{kJmo}}{{\rm{l}}^{{\rm{ - 1}}}}\).