Potassium-40 (Z=19 ,n=21) is a radioactive isotope that is rare but not unknown in nature. It is particularly interesting in that it lies along the curve of stability yet decays by both ${\mathit{\beta }}^{\mathbf{+}}$and ${\mathit{\beta }}^{\mathbf{-}}$-that is. in both directions away from the curve of stability. (a) Identify the daughter nuclei for both decays, (b) Many factors governing nuclear stability are discussed in the chapter (e.g., those in the semiempirical binding energy formula. magic numbers. and even numbers). Identify those that would argue only for ${\mathit{\beta }}^{+}$decay. (c) Which factors argue only for decay? (d) Which argue for either ${\mathit{\beta }}^{\mathbf{-}}\mathbf{}\mathbf{or}\mathbf{}{\mathit{\beta }}^{\mathbf{+}}$decay?

The given data can be listed below as:

• The number of the neutrons in the potassium-40 nucleus is N = 221.
• The potassium-40 nucleus’ atomic number is Z = 19.
• The potassium-40 nucleus’ atomic mass is A = 40.

The nuclear physics is described as the physics of the nuclear interactions and atomic nuclei. It mainly provides knowledge regarding atomic nucleus.

The equation of the nuclear reaction from the decay is expressed as:

$$\begin{gathered} {}_{19}{}^{40}K{\to }_Z^{N}X+e^{+}+V_e+Q \\ \mathrm{Here},{}_{19}{}^{40}\mathrm{K}\mathrm{is}\mathrm{the}\mathrm{potassium}-40\mathrm{nucleus}{,}_{\mathrm{Z}}^{\mathrm{N}}\mathrm{X}\mathrm{is}\mathrm{the}\mathrm{daughter}\mathrm{nucleus},{\mathrm{e}}^{+}\mathrm{is}\mathrm{the} \\ \mathrm{electron},{\mathrm{v}}_{\mathrm{e}}\mathrm{is}\mathrm{the}\mathrm{electron}\mathrm{neutrino}\mathrm{and}\mathrm{Q}\mathrm{is}\mathrm{the}\mathrm{released}\mathrm{kinetic}\mathrm{energy}. \end{gathered}$$

Here,${}_{19}{}^{40}K$ is the potassium-40 nucleus,${}_Z{}^{N}X$ is the daughter nucleus,$e^{+}$ is the electron,$V_e$ is the electron neutrino and Q is the released kinetic energy.

As the ${\beta }^{+}$proton decay has been transformed into Neutron by releasing ${\beta }^{+}$particle, it shows that the daughter nucleus may have the same mass number as the parent nucleus and also atomic number for about smaller than the nucleus that is parent nucleus. The daughter nucleus’ atomic number can be obtained by subtracting 1 from potassium’s atomic number. Hence, the daughter nucleus decays into Argon-40.

${}_{19}{}^{40}K{\to }_{18}^{40}Ar+e^{+}+V_e+Q$

As the${\beta }^{-}$ decay is observed into the potassium-40 nucleus, then the daughter nucleus will decay into the helium atom.

${}_{19}{}^{40}K{\to }_Z^{N}X+e^{-}+{\overline{V}}_e+Q$

Here,${\overline{v}}_e$ is electron antineutrino.

As the ${\beta }^{-}$proton decay has been transformed into Neutron by releasing ${\beta }^{-}$particle, it shows that the daughter nucleus may have the same mass number as the parent nucleus and also atomic number for about smaller than the nucleus that is parent nucleus. The daughter nucleus’ atomic number can be obtained by adding from potassium’s atomic number. Hence, the daughter nucleus decays into Calcium-40.

${}_{19}{}^{40}K{\to }_Z^{N}Ca+e^{-}+{\overline{V}}_e+Q$

Thus, the daughter nuclei for both decays are Argon-40 and Calcium-40 nucleus.

The Nucleus mainly tends to lower the binding energy with the help of releasing particles by radioactive decay process. In the formula of semiempirical binding energy, the asymmetry term is $-c_4.\frac{{\left(N-Z\right)}^2}{A}$which significantly decreases the nucleus’ binding energy by decreasing the atomic number.

Thus, the asymmetry term that would argue only for${\beta }^{-}$ decay.

The Nucleus mainly tends to lower the binding energy with the help of releasing particles by radioactive decay process. In the${\beta }^{+}$ decay process, the neutron is mainly transformed into the proton by releasing positron which shows that the proton number decreases. In the formula of semiempirical binding energy, the coulomb term is$-c_3.\frac{Z\left(Z-1\right)}{A^{1/3}}$ which significantly decreases the nucleus’ binding energy by increasing the atomic number.

Thus, the coulomb term that would argue only for${\beta }^{+}$ decay.

The nucleus mainly goes into radioactive decay for producing a nucleus’ stable configuration. In ${\beta }^{+}\mathrm{and}{\beta }^{-}$, the proton and the neutron’s decay number are changed. Inside a stable nucleus, the neutron and the proton number get equal to a magic number. That shows that a nucleus changes until it has a magic neutron and proton number.

Thus, the magic number argue for either ${\beta }^{+}\mathrm{and}{\beta }^{-}$decay.