Chapter 24, Problem 81A

Interpretation Introduction

Interpretation:

The product nucleus in of bromine-80 in gamma, positron and electron emission needs to be determined.

Concept introduction:

Gamma particle is just a photon.

Positron emission takes place when a proton inside the nucleus is converted into a neutron and positron and neutrino are emitted.

Electron capture is when an electron falls into the nucleus.

Answer to Problem 81A

Gamma emission, ${}_{35}^{80}\text{Br}{\to }_{\text{35}}^{\text{80}}\text{Br}{+}_{\text{0}}^{\text{0}}\text{γ}$

Positron emission, ${}_{\text{35}}^{\text{80}}\text{Br}{\to }_{\text{34}}^{\text{80}}\text{Br}{+}_{\text{1}}^{\text{0}}\text{e}{+}_{\text{0}}^{\text{0}}\text{v}$

Electron capture, ${}_{\text{35}}^{\text{80}}{\text{Br+}}_{-1}^{0}e{\to }_{34}^{80}Se{+}_0^{0}v$

Explanation of Solution

Gamma particle Which means it has no mass and no charge. As a result, the nucleus that emits the gamma particle will remain unchanged. Therefore, ${}_{35}^{80}\text{Br}{\to }_{\text{35}}^{\text{80}}\text{Br}{+}_{\text{0}}^{\text{0}}\text{γ}$

Positron emission Positron have same mass but positive charge. Since, the proton is converted into a neutron, so the atomic number will decrease by 1 and atomic mass will remain same. Therefore, ${}_{\text{35}}^{\text{80}}\text{Br}{\to }_{\text{34}}^{\text{80}}\text{Br}{+}_{\text{1}}^{\text{0}}\text{e}{+}_{\text{0}}^{\text{0}}\text{v}$

Electron capture Since the proton is converted into neutron and neutrino is emitted. Since, the proton is converted into a neutron, so the atomic number will decrease by 1 and atomic mass will remain same. Therefore, ${}_{\text{35}}^{\text{80}}{\text{Br+}}_{-1}^{0}e{\to }_{34}^{80}Se{+}_0^{0}v$

Conclusion

Br-80 decays into Se-80 in both positron emission and electron capture.