Question

A rare decay mode has been observed in which²²²Ra emits a¹⁴C nucleus. (a) The decay equation is\(^{222}Ra{ \to ^A}X{ + ^{14}}C\). Identify the nuclide^AX. (b) Find the energy emitted in the decay. The mass of²²²Ra is222.015353 µ.

Short answer

(a) The nuclide of ^AX is obtained as: \(_Z^A{X_N} = _{82}^{208}P{b_{126}}\).

(b) The energy emitted in decay is obtained as: E = 33.0683 MeV.

Step-by-step solution

Define Radioactivity

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.

Identifying the nuclide of AX

a. The value of ²²²Ra emits the value of ¹⁴C.

With the help of the decay equation: A= 222 - 14= 208

Z=88 - 6 = 82

N = 134- 8 = 126

Then, the nuclide we will have is:\(_Z^A{X_N}{\rm{ }} = {\rm{ }}_{82}^{208}P{b_{126}}\).

Therefore, the nuclide is: \(_Z^A{X_N}{\rm{ }} = {\rm{ }}_{82}^{208}P{b_{126}}\).

Evaluating the energy emitted in decay

To evaluate the changing in mass which leads to the emitted energy because of the decay, the following relation is used:

\(\begin{align}{}\Delta m{\rm{ }} &= {\rm{ }}{m_{Ra}} - {m_{Pb}} - {m_C}\\ &= {\rm{ }}222.0153\,{\rm{a}}{\rm{.m}}{\rm{.u}} - 207.9766\,{\rm{a}}{\rm{.m}}{\rm{.u}} - 14.0032\,{\rm{a}}{\rm{.m}}{\rm{.u}}\\ &= {\rm{ }}0.0355\,{\rm{a}}{\rm{.m}}{\rm{.u}}\end{align}\)

Then, according to Einstein mass/energy equation, we obtain:

\(\begin{align}{}E{\rm{ }} &= {\rm{ }}\Delta m{c^2}\\ &= {\rm{ }}0.0355\,{\rm{a}}{\rm{.m}}{\rm{.u }} \times {\rm{ }}931.5\,{\rm{MeV/u}}{{\rm{c}}^{\rm{2}}}{\rm{ }} \times {\rm{ }}{c^2}\\ &= {\rm{ }}33.0683\,{\rm{MeV}}\end{align}\)

Therefore, the energy emitted is: 33.0683 MeV.