Question

Approximately what is the total energy of the neutrino emitted when ${ }_{11}^{22} \mathrm{Na}$ decays by electron capture?

Short answer

Answer: The approximate total energy of the emitted neutrino is 2.85 MeV.

Step-by-step solution

Identify the daughter nucleus

After electron capture, a proton in the parent nucleus transforms into a neutron, thus decreasing the atomic number (Z) by 1 while keeping the mass number (A) constant. So, the daughter nucleus will have an atomic number of 10 and a mass number of 22, which is \({}_{10}^{22}\mathrm{Ne}\).

Calculate the mass difference between parent and daughter nuclei

We can find the mass difference by using the atomic mass values for the parent (\({m}_{\mathrm{Na}}\)) and daughter (\({m}_{\mathrm{Ne}}\)) nuclei. Using a reference database, we find the atomic mass values as follows: \({m}_{\mathrm{Na}} = 21.9944364\) amu \({m}_{\mathrm{Ne}} = 21.9913849\) amu The mass difference (\(Δm\)) is obtained by subtracting the daughter nucleus mass from the parent nucleus mass: \(Δm = {m}_{\mathrm{Na}} - {m}_{\mathrm{Ne}} = 21.9944364 - 21.9913849 = 0.0030515\) amu

Convert the mass difference to energy

Now, we'll use the energy-mass equivalence principle (E=mc²) to convert the mass difference into energy. First, convert the mass difference from atomic mass units (amu) to kilograms (kg), using the conversion factor 1 amu = 1.66054 × 10^{-27} kg: \(Δm_\mathrm{kg} = 0.0030515 \times 1.66054 × 10^{-27} = 5.06895 × 10^{-30}\) kg Next, apply the energy-mass equivalence principle with the speed of light (c) value as 3 × 10^8 m/s: \(E = Δm_\mathrm{kg} \times c^2 = 5.06895 × 10^{-30} \times (3 × 10^8)^2 = 4.5620 × 10^{-13}\) Joules

Express the energy in a suitable unit and approximate the value

Nuclear reactions usually deal with energy in the unit of electronvolts (eV). So, we'll convert the calculated energy into MeV (Mega-electronvolts), using the conversion factor 1 Joule = 6.242 × 10^{12} eV: \(E_\mathrm{MeV} = 4.5620 × 10^{-13} \times 6.242 × 10^{12} \times 10^{-6} = 2.847\) MeV Approximately, the total energy of the emitted neutrino in the electron capture decay of \({}_{11}^{22}\mathrm{Na}\) is 2.85 MeV.