# Question #cbe87

May 8, 2015

Here's how the nuclear equations would look for type of decay.

• Gamma emission

A gamma particle is just a photon, which in turn is a particle that has no mass and no charge. As a result, the nuleus that emits the gamma particle will remain unchanged.

In your case, the nuclear equation will look like this

${\text{_35^80"Br" -> ""_35^80"Br" + }}_{0}^{0} \gamma$

• Positron emission

Positron decay takes place when a proton inside the nucleus is converted into a neutron; at the same time, a positron and a neutrino are emitted.

A positron is the antiparticle of an electron, i.e. same mass, but a positive charge. Since a proton is converted into a neutron, the atomic number will decrease by 1, but the atomic mass will remain unchanged.

${\text{_35^80"Br" -> ""_34^80"Se" + ""_1^0e + }}_{0}^{0} \nu$

Notice that Br-80 decays into Se-80.

Electron capture takes place when an electron falls into the nucleus. As a result, a proton is converted into a neutron and a neutrino is emitted.

Once again, the conversion of a proton into a neutron will decrease the atomic number by 1, but keep the atomic mass unchanged.

${\text{_35^80"Br" + ""_text(-1)^0e -> ""_34^80"Se" + }}_{0}^{0} \nu$

Once again, Br-80 will decay into Se-80.