Objectives:

• To understand beta-minus decay and beta-plus decay in terms of both the particles emitted from the nucleus and the quark model;
(i) β- decay in terms of a quark model; $d \rightarrow u + \ ^{0}_{-1}e + \bar{\nu_{e}}$
(ii) β+ decay in terms of a quark model; $u \rightarrow d + \ ^{0}_{+1}e + \nu_{e}$;
• To understand the balancing of quark transformation equations in terms of charge
• To acknowledge the decay of particles in terms of the quark model

Beta Decay

There are three different types of beta decay processes; beta-minus ($\beta^{-}$) decay, orbital electron capture, and beta-plus ($\beta^{+}$) decay – otherwise known as positron emission. You will only be required to understand the $\beta^{-}$ and $\beta^{+}$ processes, click here if you would like to learn a little more about orbital electron capture.

Beta particles can be easily distinguishable from alpha particles due to their substantially greater range in air.

Beta-Minus (Negatron) Emission

The section on beta emission on the previous page (radioactive decay and nuclear equations) focussed predominantly on beta-minus emission. Beta-minus decay occurs when an electron (negatively charged – hence beta-‘minus’) is ejected from the nucleus of a radioactive element. During any type of nuclear decay conservation rules must apply; during GCSE and on the previous page, both the atomic mass and atomic number are  conserved such that the following example regarding Thorium-234 (which undergoes $\beta^{-}$ emission) can be written;

$^{234}_{90}Th \ \ \rightarrow \ \ ^{234}_{91}Pa + \ ^{0}_{-1}e$

During this decay, the product has same mass number as the original nuclei, but its atomic number is greater by one unit; the charge has been conserved.

If we look at this decay more closely you can see that it is a neutron that has changed into a proton and an electron. This is evident due to the fact that the mass number is the same but the atomic number has increased by one unit. We can therefore write the following nuclear decay;

$^{1}_{0}n \ \ \rightarrow \ \ ^{1}_{1}p + \ ^{0}_{-1}e$

Since we know that a neutron is made up of an up quark and two down quarks and a proton is made up of two up quarks and a down quark, we can write the following quark equation;

$udd \ \ \rightarrow \ \ uud + \ ^{0}_{-1}e$

From this you can see that one the original up quark remains the same, as does one of the down quarks. However one of the down quarks has transformed into an up quark, so the following quark transformation equation can be written;

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

In terms of charge this equation works;  $-\frac{1}{3} \ \rightarrow \ +\frac{2}{3} + -1 = -\frac{1}{3}$

However, after learning about antimatter, baryons and leptons it can be shown that other quantities are not conserved;

Baryon number:

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

Baryon number has been conserved. ✓

Lepton number:

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

Lepton number has not been conserved. The equation needs to balance lepton number some how. What particle has zero charge (charge is already conserved so we don’t want to change that), and has a lepton number of $-1$ ?

A neutrino has no charge, however they have a lepton number of +1.
An antineutrino has no charge and has a lepton number of -1.
An antineutrino fits the criteria we want.

$d \ \ \rightarrow \ \ u + \ ^{0}_{-1}e + \ \bar{\nu}$

If an electron is emitted an antielectron-neutrino is emitted (as a rule of thumb, the type of neutrino that is emitted is in the same generation of fermions as the charged lepton (electron, muon or tau). This quark transformation therefore becomes;

$d \ \ \rightarrow \ \ u + \ ^{0}_{-1}e + \ \bar{\nu_{e}}$

Quantities conserved:

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

This quark transformation tells us that when a down quark transforms into an up quark, an electron and antielectron-neutrino are emitted. Therefore our initial decay equation involving Thorium-234 is incorrect as it does not show the neutrino being emitted. The correct equation is;

$^{234}_{90}Th \ \ \rightarrow \ \ ^{234}_{91}Pa + \ ^{0}_{-1}e + \ \nu_{e}$

If you want more information on $\beta^{-}$ decay try this YouTube video:

Beta-Plus (Positron) Emission

During $\beta^{+}$ emission, instead of an electron being emitted from the nucleus, a positron is instead (positively charged – hence beta-‘plus’). Since a positron (an anti-electron) is emitted the mass number stays the same but the atomic number decreases by one unit;

$u \ \ \rightarrow \ \ d + \ ^{0}_{+1}e$

Following the same process as for $\beta^{-}$, the charge baryon and lepton numbers should be checked to see if they have been conserved;

Charge:

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

Charge is conserved. ✓

Baryon number:

$u \ \ \rightarrow \ \ d + \ ^{0}_{+1}e$
$+\frac{1}{3} \ \rightarrow \ +\frac{1}{3} + \ 0$
$+\frac{1}{3} \ \rightarrow \ +\frac{1}{3}$

Baryon number is conserved. ✓

Lepton number:

$u \ \ \rightarrow \ \ d + \ ^{0}_{+1}e$
$0 \ \rightarrow \ 0 + \ +1$
$0 \ \rightarrow \ +1$

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 electron-neutrino.

Therefore the full quark transformation becomes:

$u \ \ \rightarrow \ \ d + \ ^{0}_{+1}e + \ {\nu_{e}}$

Quantities conserved:

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

If you want more information on $\beta^{+}$ decay try this YouTube video:

What determines whether the decay is $\beta^{-}$ or $\beta^{+}$ ?

You should familiarise yourself with the Segrè plot;

The Segrè plot shows neutron number against proton number. The blue line on the graph represents the most stable nuclei, the grey dots surrounding the blue line shows the isotopes that have been discovered.

• An isotope on the left hand side of the blue line has an excess number of neutrons, or not enough protons for it to be stable, it would therefore likely undergo $\beta^{-}$ decay – to transform a neutron into a proton, emitting an electron and antielectron-neutrino. For a neutron to transform into a neutron, a down quark needs to transform into an up quark (as outlined in the above examples);  $d \ \ \rightarrow \ \ u + \ ^{0}_{-1}e + \ \bar{\nu_{e}}$
• An isotope on the right hand side of the blue line has an excess number of protons, or not enough neutrons for it to be stable, it would therefore likely undergo $\beta^{+}$ decay – to transform a proton into a neutron, emitting a positron and a electron-neutrino. For a proton to transform into a neutron, an up quark needs to transform into a down quark (as outlined in the above examples);  $u \ \ \rightarrow \ \ d + \ ^{0}_{+1}e + \ {\nu_{e}}$