 The three types of beta emission are;

• Beta-minus ( $\beta^{-}$) decay
• Beta-plus ( $\beta^{-}$) decay – otherwise known as positron emission
• Orbital electron capture  $\beta^{-}$ and $\beta^{+}$ emission are discussed here. Orbital electron capture is a process that can occur when the ratio of neutrons to protons is low.

In this process, instead of a proton being converted into a neutron with the emission of a positron (and electron-neutrino), a proton rich nucleus of an electrically neutral atom absorbs an electron from one of the inner electron shells (or energy levels). This process there converts a proton into a neutron; $^{0}_{-1}e + \ ^{1}_{1}p \ \ \rightarrow \ \ ^{1}_{0}n$

As shown on the page focusing on $\beta^{-}$ and $\beta^{+}$, nuclear equations need to ensure charge, baryon number and lepton number are conserved, so first lets write a quark equation; $^{0}_{-1}e + \ ^{1}_{1}p \ \ \rightarrow \ \ ^{1}_{0}n$ $^{0}_{-1}e + uud \ \ \rightarrow \ \ ^udd$ $^{0}_{-1}e + u \ \ \rightarrow \ \ d$

Charge: $^{0}_{-1}e + u \ \ \rightarrow \ \ d$ $-1 + \frac{2}{3} \ \ \rightarrow \ \ -\frac{1}{3}$ $-\frac{1}{3} \ \ \rightarrow \ \ -\frac{1}{3}$

Charge is conserved. ✓

Baryon number: $^{0}_{-1}e + u \ \ \rightarrow \ \ d$ $0 + \frac{1}{3} \ \ \rightarrow \ \ \frac{1}{3}$ $\frac{1}{3} \ \ \rightarrow \ \ \frac{1}{3}$

Baryon number is conserved. ✓

Lepton number: $^{0}_{-1}e + u \ \ \rightarrow \ \ d$ $+1 + 0 \ \ \rightarrow \ \ + 0$ $+1 \ \ \rightarrow \ \ 0$

Lepton number is not conserved. ✓ So an additional particle must be emitted that does not affect the charge or the baryon number but does change the lepton number. What particle has no charge, no baryon number and a lepton number of -1?

It must be a lepton and for no charge to be present the particle must be one of neutrinos – an antielectron-neutrino.

And so, the full quark transformation for orbital electron capture must be: $^{0}_{-1}e + u \ \ \rightarrow \ \ d + \ \bar{\nu{e}}$

Quantities conserved:

• Charge: ✓
• Baryon number: ✓
• Lepton number: ✓

In addition to this the full process in which a proton transforms into a neutron is: $^{0}_{-1}e + \ ^{1}_{1}p \ \ \rightarrow \ \ ^{1}_{0}n + \ \bar{\nu_{e}}$ 