Question

(a) Write the complete reaction equation for electron capture by\({}^{{\rm{15}}}{\rm{O}}\). (b) Calculate the energy released.

Short answer

(a) The equation is\({}_8^{15}{O_7} + {e^ - } \to {}_7^{15}{N_8} + {v_e}\).

(b) The energy released is \(2.7527\,{\rm{MeV}}\).

Step-by-step solution

Define energy

Energy is a measurable attribute that may be transferred from one thing to another in order for it to do work.

Explanation

(a) The full electron capture reaction of \({}^{{\rm{15}}}{\rm{O}}\) is,

\({}_8^{15}{O_7} + {e^ - } \to {}_7^{15}{N_8} + {v_e}\)

Evaluating the energy released

(b) The isotopes' masses are,

\(\begin{array}{l}m({}^{15}O) = 15.0030656\,{\rm{u}}\\m({}^{15}N) = 15.0001089\,{\rm{u}}\end{array}\)

As a result, the mass difference is,

\(\begin{array}{c}\Delta m = \left( {15.0030656\,{\rm{u}}} \right) - \left( {15.0001089\,{\rm{u}}} \right)\\ = 0.0029567\,{\rm{u}}\end{array}\)

We can calculate the energy

released by the process,

\(\begin{array}{c}E = \Delta m{c^2}\\E = (0.0029567\,{\rm{u}})(931.5\,{\rm{MeV/u}}{{\rm{c}}^2}) \times {c^2}\\ = 2.7527\,{\rm{MeV}}\end{array}\)

Therefore, the energy released is\(2.7527\,{\rm{MeV}}\).