Question

Both \(\beta^{-}\) and \(\beta^{+}\) emissions are observed for artificially produced radioisotopes of low atomic numbers, but only \(\beta^{-}\) emission is observed with naturally occurring radioisotopes of high atomic number. Why do you suppose this is so?

Short answer

\(\beta^{+}\) emission is less observed in nature because it requires a lower neutron-to-proton ratio and is less energetically favourable. High atomic number elements prefer \(\beta^{-}\) decay to increase stability by decreasing their proton count.

Step-by-step solution

Understanding the Difference between \(\beta^{-}\) and \(\beta^{+}\) decay

In \(\beta^{-}\) decay, a neutron inside the nucleus transforms into a proton and emits an electron (and an electron antineutrino). This increases the atomic number by 1. \(\beta^{+}\) decay is the opposite, where a proton inside the nucleus changes into a neutron and emits a positron (and an electron neutrino), thereby decreasing the atomic number by 1.

Connection to Atomic number

For elements with high atomic numbers, they have a high proton count which can cause instability due to repulsion between protons. To reduce this repulsion and increase the stability of the nucleus, it prefers the transformation process that allows it to reduce its proton count i.e., \(\beta^{-}\) decay. Conversely, lighter or artificially produced elements with low atomic numbers might have an excess of neutrons, leading to \(\beta^{+}\) decay to increase their proton count.

Understanding why \(\beta^{+}\) decay rarely occurs in nature

In nature, isotopes that undergo \(\beta^{+}\) decay are rare because it requires the atom to have a lower neutron-to-proton ratio. Most naturally occurring isotopes have a higher neutron-to-proton ratio which favours \(\beta^{-}\) decay. Additionally, \(\beta^{+}\) decay is less energetically favourable because it requires the nucleus to possess sufficient energy to convert a proton to a neutron, which is a heavier particle.

Key concepts

Beta Decay

Beta decay is a process by which unstable nuclei transform to become more stable. There are primarily two types of beta decay: \( \beta^{-} \) decay and \( \beta^{+} \) decay. In \( \beta^{-} \) decay:

• A neutron in the nucleus transforms into a proton.
• An electron and an antineutrino are emitted as by-products.

This transformation increases the atomic number of the element by 1 unit.On the contrary, during \( \beta^{+} \) decay:

• A proton in the nucleus is converted into a neutron.
• A positron and a neutrino are released in this process.

The atomic number of the element decreases by 1. \( \beta^{-} \) decay is more common because it allows for the reduction in neutron-heavy situations, assisting in achieving a balanced state. It is energetically more favourable, especially for naturally occurring isotopes.

Neutron-to-Proton Ratio

The neutron-to-proton ratio within a nucleus is a crucial factor that determines nuclear stability. It informs whether an atom is more likely to undergo beta decay.High neutron-to-proton ratio scenarios:

• Typically, atoms with a high number of neutrons compared to protons will experience \( \beta^{-} \) decay.
• This process converts a neutron to a proton, addressing the excess in neutrons and moving towards stability.

Low neutron-to-proton ratio circumstances:

• An excess of protons can lead to \( \beta^{+} \) decay, which increases the neutron count and balances the ratio.
• However, naturally occurring isotopes rarely have this condition, making \( \beta^{+} \) decay uncommon.

Thus, the neutron-to-proton ratio effectively determines the type of beta decay an isotope might undergo to achieve nuclear stability.

Nuclear Stability

Nuclear stability relates to the ability of a nucleus to remain unchanged over time. For an atom to be stable, the forces within the nucleus must be balanced, particularly the interactions between protons and neutrons.Factors influencing nuclear stability include:

• The neutron-to-proton ratio, which must fall within a range that promotes stability.
• Strong nuclear forces that keep the protons and neutrons bound together.

Why is \( \beta^{-} \) decay prevalent?

• In isotopes of high atomic numbers, there is often excessive protonic repulsion.
• This makes \( \beta^{-} \) decay more favourable as it converts neutrons to protons, lessening repulsive forces and enhancing stability.

On the other hand, \( \beta^{+} \) decay is rare. It needs exceptionally high energy levels because converting a neutron to a proton (or the reverse for \( \beta^{+} \)) is challenging and less energetically favourable in nature. Hence, natural radioisotopes rarely undergo \( \beta^{+} \) decay.