### Writing Positron Decay and Electron Capture Equations

Positron Decay

Positron decay is like a mirror image of beta decay. These points present a simplified view of what positron decay actually is:

1) Something inside the nucleus of an atom breaks down, which causes a proton to become a neutron.
2) It emits a positron and a neutrino which go zooming off into space.
3) The atomic number goes DOWN by one and mass number remains unchanged.

Example #1: A positron decay equation:

 ${\text{}}_{12}^{23}\text{Mg}$ ---> ${\text{}}_{11}^{23}\text{Na}$ + +  $\mathrm{\nu e}$

1) The nuclide that decays is the one on the left-hand side of the equation.
2) The order of the nuclides on the right-hand side can be in any order.
3) The way it is written above is the usual way.
4) The mass number and atomic number of the neutrino are zero.
5) The neutrino symbol is the Greek letter "nu."

Example #2: Here is another example of a positron decay equation:

 ${\text{}}_{25}^{50}\text{Mn}$ ---> ${\text{}}_{24}^{50}\text{Cr}$ + +  $\mathrm{\nu e}$

Notice that all the atomic numbers on both sides ADD UP TO THE SAME VALUE and the same for the mass numbers.

By the way, an older style for the neutrino symbol adds on two zeros where the atomic number and the mass number are placed, as well as dropping the subscripted e. I couldn't make the formatting work, so I have to describe it in words. You might wind up with an older teacher who insists on the older style of writing the neutrino. Or, you might be using an older set of materials.

Example #3: Write out the full positron decay equation for these five.

 ${\text{}}_{35}^{75}\text{Br}$ ${\text{}}_{15}^{30}\text{P}$ ${\text{}}_{14}^{27}\text{Si}$ ${\text{}}_{22}^{45}\text{Ti}$ ${\text{}}_{5}^{8}\text{B}$

Solution:

 ${\text{}}_{35}^{75}\text{Br}$ ---> ${\text{}}_{34}^{75}\text{Se}$ + +  $\mathrm{\nu e}$ ${\text{}}_{15}^{30}\text{P}$ ---> ${\text{}}_{14}^{30}\text{Si}$ + +  $\mathrm{\nu e}$ ${\text{}}_{14}^{27}\text{Si}$ ---> ${\text{}}_{13}^{27}\text{Al}$ + +  $\mathrm{\nu e}$ ${\text{}}_{22}^{45}\text{Ti}$ ---> ${\text{}}_{21}^{45}\text{Sc}$ + +  $\mathrm{\nu e}$ ${\text{}}_{5}^{8}\text{B}$ ---> ${\text{}}_{4}^{8}\text{Be}$ + +  $\mathrm{\nu e}$

Example #4: Here are five more to work on. Sometimes, the teacher wants the neutrino left off the answer. That's what I did with the answers.

 ${\text{}}_{19}^{37}\text{K}$ ${\text{}}_{27}^{54}\text{Co}$ ${\text{}}_{30}^{61}\text{Zn}$ ${\text{}}_{31}^{68}\text{Ga}$

Solution:

 ---> + ${\text{}}_{19}^{37}\text{K}$ ---> ${\text{}}_{18}^{37}\text{Ar}$ + ${\text{}}_{27}^{54}\text{Co}$ ---> ${\text{}}_{26}^{54}\text{Fe}$ + ${\text{}}_{30}^{61}\text{Zn}$ ---> ${\text{}}_{29}^{61}\text{Cu}$ + ${\text{}}_{31}^{68}\text{Ga}$ ---> ${\text{}}_{30}^{68}\text{Zn}$ +

Example #5: And here are five more. I'll only show the daughter nuclide:

 ${\text{}}_{20}^{39}\text{Ca}$ ${\text{}}_{38}^{83}\text{Sr}$

Solution:

 ---> ---> ${\text{}}_{20}^{39}\text{Ca}$ ---> ${\text{}}_{19}^{39}\text{K}$ ${\text{}}_{38}^{83}\text{Sr}$ ---> ${\text{}}_{37}^{83}\text{Rb}$ --->

Bonus Example: Five more, but no answers.

 ${\text{}}_{11}^{21}\text{Na}$ ${\text{}}_{13}^{25}\text{Al}$ ${\text{}}_{16}^{30}\text{S}$ ${\text{}}_{21}^{42}\text{Sc}$ ${\text{}}_{29}^{60}\text{Cu}$

Electron Capture

Electron capture is not like the other three decays I have covered: alpha, beta, and position. All other decays shoot something out of the nucleus. In electron capture, something ENTERS the nucleus. These points present a simplified view of what electron capture is:

1) An electron from the closest energy level falls into the nucleus, which causes a proton to become a neutron.
2) A neutrino is emitted from the nucleus.
3) Another electron falls into the empty energy level and so on causing a cascade of electrons falling. One free electron, moving about in space, falls into the outermost empty level. (Incidently, this cascade of electrons falling creates a characteristic cascade of lines, mostly (I think) in the X-ray portion of the spectrum. This is the fingerprint of electron capture.)
4) The atomic number goes DOWN by one and mass number remains unchanged.

Example #1: Here is an example of a electron capture equation:

 ${\text{}}_{36}^{81}\text{Kr}$ + ---> ${\text{}}_{35}^{81}\text{Br}$ +  $\mathrm{\nu e}$

1) The nuclide that decays is the one on the left-hand side of the equation.
2) The electron must also be written on the left-hand side.
3) A neutrino is involved. It is ejected from the nucleus where the electron reacts, so it is written on the right-hand side.

Example #2: Here's another electron capture equation:

 + ---> +  $\mathrm{\nu e}$

Notice that all the atomic numbers on both sides ADD UP TO THE SAME VALUE and the same for the mass numbers.

Example #3: Write out the full electron capture equation for the following five nuclides.

 ${\text{}}_{18}^{38}\text{Ar}$ ${\text{}}_{38}^{80}\text{Sr}$

Solution:

 ${\text{}}_{18}^{38}\text{Ar}$ + ---> ${\text{}}_{17}^{38}\text{Cl}$ +  $\mathrm{\nu e}$ ${\text{}}_{38}^{80}\text{Sr}$ + ---> ${\text{}}_{37}^{80}\text{Rb}$ +  $\mathrm{\nu e}$ + ---> +  $\mathrm{\nu e}$ + ---> +  $\mathrm{\nu e}$ + ---> +  $\mathrm{\nu e}$

Example #4: Here are five more to work on. I left off the neutrino.

 ${\text{}}_{27}^{57}\text{Co}$ ${\text{}}_{33}^{73}\text{As}$

Solution:

 ${\text{}}_{27}^{57}\text{Ar}$ + ---> ${\text{}}_{26}^{57}\text{Fe}$ ${\text{}}_{33}^{73}\text{As}$ + ---> ${\text{}}_{32}^{73}\text{Ge}$ + ---> + ---> + --->

Example #5: And a final five, with just the daughter nuclide:

 ${\text{}}_{31}^{67}\text{Ga}$ ${\text{}}_{44}^{97}\text{Ru}$ ${\text{}}_{101}^{257}\text{Md}$

Solution:

 ${\text{}}_{31}^{67}\text{Ga}$ ---> ${\text{}}_{30}^{67}\text{Zn}$ ${\text{}}_{44}^{97}\text{Ru}$ ---> ${\text{}}_{43}^{97}\text{Tc}$ ---> ---> ${\text{}}_{101}^{257}\text{Md}$ ---> ${\text{}}_{100}^{257}\text{Fm}$

Bonus Example: Five more, but no answers.

 ${\text{}}_{23}^{50}\text{V}$ ${\text{}}_{24}^{51}\text{Cr}$ ${\text{}}_{28}^{56}\text{Ni}$ ${\text{}}_{41}^{91}\text{Nb}$

Nov. 21, 2020 − I continue to learn new things! There is something called a double electron capture. About 34 nuclei are predicted to undergo double electron capture, but only three have been observed. The Wikipedia page has more information on this rare decay.

Before heading over there, you might try writing the double electron capture equation for . The answer is on the Wiki page.