A\({}_{\rm{4}}^{\rm{7}}{\rm{Be}}\) atom (mass\({\rm{ = 7}}{\rm{.0169 amu}}\)) decays into a\({}_{\rm{3}}^{\rm{7}}{\rm{Li}}\) atom (mass\({\rm{ = 7}}{\rm{.0160 amu}}\)) by electron capture. How much energy (in millions of electron volts, \({\rm{MeV}}\)) is produced by this reaction?

Radioactive decay is known as the process of an unstable atomic nucleus losing energy through radiation. The term "radioactive" refers to a substance that contains unstable nuclei. Alpha decay (\({\rm{\alpha - }}\)decay), beta decay (\({\rm{\beta - }}\)decay), and gamma decay (\({\rm{\gamma - }}\)decay) are three of the most prevalent types of decay, all of which entail the emission of one or more particles.

The reaction is

\(_{\rm{4}}^{\rm{7}}{\rm{Be + }}_{{\rm{ - 1}}}^{\rm{0}}{\rm{e}} \to _{\rm{3}}^{\rm{7}}{\rm{Li}}\)

The energy can be obtained using the formula:

\(\begin{array}{c}{\rm{E = m}} \cdot {{\rm{c}}^{\rm{2}}}\\{\rm{m = m( reactants ) - m( products )}}\\{\rm{m = (7}}{\rm{.0169 amu + 0}}{\rm{.0005 amu) - 7}}{\rm{.0160 amu}}\\{\rm{m = 1}}{\rm{.4}} \cdot {\rm{1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{amu}}\end{array}\)

Converting\({\rm{amu}}\)in to\({\rm{kg}}\)–

\(\begin{array}{c}{\rm{1 amu = 1}}{\rm{.6605}} \cdot {\rm{1}}{{\rm{0}}^{{\rm{ - 27}}}}{\rm{\;kg}}\\{\rm{m = 1}}{\rm{.4}} \cdot {\rm{1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{amu \times 1}}{\rm{.6605}} \cdot {\rm{1}}{{\rm{0}}^{{\rm{ - 27}}}}{\rm{\;kg}} \cdot {\rm{am}}{{\rm{u}}^{{\rm{ - 1}}}}\\{\rm{m = 2}}{\rm{.325}} \cdot {\rm{1}}{{\rm{0}}^{{\rm{ - 30}}}}{\rm{\;kg}}\end{array}\)

Substituting the values and calculating:

\(\begin{array}{c}{\rm{E = m}} \cdot {{\rm{c}}^{\rm{2}}}\\{\rm{E = 2}}{\rm{.325}} \cdot {\rm{1}}{{\rm{0}}^{{\rm{ - 30}}}}{\rm{\;kg \times 2}}{\rm{.99}} \cdot {\rm{1}}{{\rm{0}}^{\rm{8}}}{\rm{\;m}}{{\rm{s}}^{{\rm{ - }}{{\rm{1}}^{\rm{2}}}}}\\{\rm{E = 2}}{\rm{.078}} \cdot {\rm{1}}{{\rm{0}}^{{\rm{ - 13}}}}{\rm{\;J}}\end{array}\)

Converting energy from\({\rm{J}}\)to\({\rm{MeV}}\):

\(\begin{array}{c}{\rm{1 MeV = 1}}{\rm{.602}} \cdot {\rm{1}}{{\rm{0}}^{{\rm{ - 13}}}}{\rm{\;J}}\\{\rm{E = 2}}{\rm{.078}} \cdot {\rm{1}}{{\rm{0}}^{{\rm{ - 1}}}}{\rm{13\;J \times }}\frac{{{\rm{1 MeV}}}}{{{\rm{1}}{\rm{.602}} \cdot {\rm{1}}{{\rm{0}}^{{\rm{ - 13}}}}{\rm{\;J}}}}\\{\rm{E = 1}}{\rm{.297 MeV}}\end{array}\)

Therefore, the value for energy is obtained as\({\rm{E = 1}}{\rm{.297 MeV}}\).