Question

The nuclide ${}_{\mathbf{5}}{}^{\mathbf{8}}\mathbf{B}$ decays by positron emission to ${}_{\mathbf{5}}{}^{\mathbf{8}}\mathbf{B}$ . What is the energy released (in MeV).

Short answer

The energy released is $21.40 \text{MeV}$.

Step-by-step solution

Positron Emission

For a nucleus having $\frac{N}{Z}<1$ , positron decay takes place to stabilize the nucleus. When a positron emission takes place, a proton is transformed to a neutron. Positron emission is also accompanied by a neutrino emission. The mass number of the daughter nuclei remains same as that of the parent nucleus, but the atomic umber of the daughter nuclei is one less than that of the parent nuclei.

${}_{\text{5}}{}^{\text{8}}\text{B}\to {}_{\text{4}}{}^{\text{8}}\text{Be +}{}_{\text{+ 1}}{}^{\text{0}}{\text{e}}^{+}\text{+}\nu$

  Energy released

To calculate the kinetic energy involved in the positron decay we have to calculate the mass difference.

$$\begin{gathered} \Delta m=m\left({}_4{}^{10}B\right)+2m\left({}_1{}^0{e}^{+}\right)-m\left({}_5{}^{10}B\right) \\ =8.000530 u+2\times 0.00054858 u-8.024607u \\ =-0.022981 u \end{gathered}$$

$$\begin{gathered} \Delta E=-0.022981\times 931.5 \text{MeV} \\ =-21.40 \text{MeV} \end{gathered}$$

Therefore, the energy released is $21.40 \text{MeV}$.