Calculate the mass defect for the ${\mathit{\beta }}^{\mathbf{+}}$decay of ${}_{\mathbf{6}}{}^{\mathbf{11}}\mathbf{}$C. Is this decay energetically possible? Why or why not? The atomic mass of ${}_{\mathbf{6}}{}^{\mathbf{11}}\mathbf{}$C is 11.011434 u.

${\mathit{\beta }}^{\mathbf{+}}$decay can occur whenever the mass of the original neutral atom is at least two electron masses larger than that of the final atom.

The decay equation is given as:

${}_{\mathbf{6}}{}^{\mathbf{11}}\mathit{C}\mathbf{\to }{}_{\mathbf{Z}}{}^{\mathbf{A}}\mathit{X}\mathbf{+}{}_{\mathbf{1}}{}^{\mathbf{0}}\mathit{e}$

From the conservation of atomic number and the number of nucleons,

$$\begin{gathered} \mathbf{}\mathbf{6}\mathbf{=}\mathit{Z}\mathbf{+}\mathbf{1}\mathbf{\Rightarrow }\mathit{Z}\mathbf{=}\mathbf{5} \\ \mathbf{11}\mathbf{=}\mathit{A}\mathbf{+}\mathbf{0}\mathbf{\Rightarrow }\mathit{A}\mathbf{=}\mathbf{11} \end{gathered}$$

So, the daughter nucleus is ${}_{\mathbf{5}}{}^{\mathbf{11}}\mathit{B}$ with atomic mass of 11.009305u.

${\mathit{\beta }}^{\mathbf{+}}$decay can occur whenever the mass of the original neutral atom is at least two electron masses larger than that of the final atom.

Mass defect can be given as:

$\mathbf{∆}\mathit{M}\mathbf{=}\mathit{M}\mathbf{\left(}{}_{\mathbf{6}}{}^{\mathbf{11}}\mathit{C}\mathbf{\right)}\mathbf{-}\mathbf{2}{\mathit{m}}_{\mathbf{e}}\mathbf{-}\mathit{M}\mathbf{\left(}{}_{\mathbf{5}}{}^{\mathbf{11}}\mathit{B}\mathbf{\right)}$

Plug the values,

$\mathbf{∆}\mathit{M}\mathbf{=}\mathbf{11}\mathbf{.}\mathbf{011434}\mathbf{-}\mathbf{2}\mathbf{*}\mathbf{0}\mathbf{.}\mathbf{000548580}\mathbf{-}\mathbf{11}\mathbf{.}\mathbf{009305}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{032}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\mathit{u}$

Since the mass defect is positive, the decay is energetically possible.

Thus, this decay is energetically possible.