(a) Write the complete \({{\rm{\beta }}^{\rm{ - }}}\) decay equation for the neutron.

(b) Find the energy released in the decay.

In nuclear physics, beta decay (\({\rm{\beta }}\)-decay) is a type of radioactive decay in which an atomic nucleus emits a beta particle (fast energetic electron or positron) that transforms the original nuclide into an isobar of that nuclide.

(a)

Since\({{\rm{\beta }}^{\rm{ - }}}\)decay increases the atomic number of the nucleus, after beta decaythe neutron will become proton. The complete beta decay equation ofneutron is –

\(_0^1{n_1} \to _1^1{p_0} + {\beta ^ - }\)

Therefore, the decay reaction is \(_0^1{n_1} \to _1^1{p_0} + {\beta ^ - }\).

(b)

The mass of neutron is: \({m_n} = 1.008665\,{\rm{u}}\).

The mass of proton is: \({m_p} = 1.007276\,{\rm{u}}\).

And the mass of electron is:\({m_e} = 0.00054858\,{\rm{u}}\).

Hence, the mass difference is –

\(\begin{array}{}\delta m = (1.008665\,{\rm{u}}) - \{ (1.007276\,{\rm{u}}) + (0.00054858\,{\rm{u}})\} \\ = 0.00084042\,{\rm{u}}\end{array}\)

Hence, the energy released in the decay reaction is –

\(\begin{aligned}{}E & = \Delta m{c^2}\\E &=(0.00084042\,{\rm{u}})(931.5\,{{{\rm{MeV}}} \mathord{\left/ {\vphantom {{{\rm{MeV}}} {{\rm{u}}{{\rm{c}}^{\rm{2}}}}}} \right.}{{\rm{u}}{{\rm{c}}^{\rm{2}}}}})\times{c^2}\\ &= 0.78285\,{\rm{MeV}}\end{aligned}\)

Therefore, the value for energy is obtained as \(E = 0.78285\,{\rm{MeV}}\).