(a) Write the completeα decay equation for²²⁶Ra. (b) Find the energy released in the decay.

The spontaneous emission of radiation in the form of particles or high-energy photons as a result of a nuclear process is known as radioactivity.

\[_{88}^{226}R{a_{138}} \to _{86}^{222}R{n_{136}}{\rm{ }} + {\rm{ }}_2^4H{e_2}\]

Therefore, the decay equation is: \[_{88}^{226}R{a_{138}} \to _{86}^{222}R{n_{136}}{\rm{ }} + {\rm{ }}_2^4H{e_2}\].

The mass of ²²²Rn we have is 222.0175777 u

The mass of α particle we have is: 4.00260325415

So, the mass difference is then obtained as:

\[\begin{array}{c}\delta m = (226.0254098\,{\rm{u}}) - [(222.015353\,{\rm{u}}) + (4.00260325415\,{\rm{u}})\\ = 0.0052288\,{\rm{u}}\end{array}\]

So, the energy released in the reaction is obtained as:

\[\begin{array}{c}E = \Delta m{c^2}\\E = (0.0052288\,{\rm{u}}) \times (931.5\,{\rm{MeV/u}}{{\rm{c}}^2}) \times {c^2}\\ = 4.8707\,{\rm{MeV}}\end{array}\]

Therefore, the energy released is: 4.8707 MeV