Question

Rutherford theorized that a compound nucleus forms when helium nuclei hit nitrogen- 14 nuclei. If this compound nucleus decayed by the loss of a neutron instead of a proton, what would be the other product?

Short answer

The other product would be Oxygen-17, which has the same number of protons as nitrogen but one less neutron.

Step-by-step solution

Understand the Nuclear Reaction

First, understand that helium nuclei (alpha particles) have 2 protons and 2 neutrons. When an alpha particle hits a nitrogen-14 nucleus, which has 7 protons and 7 neutrons, they form a compound nucleus.

Determine the Composition of the Compound Nucleus

Add the protons and neutrons of the helium nucleus (alpha particle) to those of the nitrogen-14 nucleus to find the total number of protons and neutrons in the compound nucleus.

Calculate the Loss of a Neutron

After the formation of the compound nucleus, if it loses a neutron, then the number of protons remains the same while the number of neutrons decreases by one.

Identify the Resulting Nucleus

Given that there is no change in the number of protons but there is a loss of one neutron, look up the element that corresponds to the remaining number of protons and the new number of neutrons to determine the other product.

Key concepts

Alpha Particles

Alpha particles play an essential role in nuclear reactions and serve as one of the paths for nuclear decay. They consist of two protons and two neutrons, and are thus helium-4 nuclei. Represented by the symbol \( \alpha \), alpha particles are relatively large and carry a double-positive charge due to their composition.

When studying these particles, it is crucial to note their influence on nuclear reactions such as in the famous Rutherford's experiment. Upon colliding with other nuclei, alpha particles can trigger a transformation process, leading to the creation of a different element or isotope. For instance, when an alpha particle interacts with a nitrogen-14 nucleus, the combination leads to the formation of a compound nucleus, potentially altering the nuclear structure and setting the stage for further decay.

Nuclear Decay

Nuclear decay is a spontaneous process through which an unstable atomic nucleus releases energy and particles to reach a more stable state. Among the varieties of nuclear decay, we can identify alpha decay, beta decay, and gamma decay, each involving the emission of different particles or electromagnetic radiation.

In the context of our exercise, when the compound nucleus formed from an alpha particle and a nitrogen-14 nucleus loses a neutron, it is partaking in a specific type of decay process. The loss of a neutron does not alter the atomic number (which determines the element), but it does change the neutron count, thereby producing a different isotope of the original element. This transformation illustrates one aspect of nuclear decay that can result in an array of isotopes, some of which could be radioactive, further continuing the decay chain.

Compound Nucleus

A compound nucleus is a short-lived, intermediate state formed when two atomic nuclei fuse together during a nuclear reaction. The process involves a significant amount of energy and may lead to the creation of entirely new elements or isotopes, depending on the particles involved in the subsequent decay.

In our example from the exercise, the collision of alpha particles with nitrogen-14 nuclei results in a compound nucleus that temporarily contains the combined protons and neutrons of both participating entities. This formation then undergoes a decay process, where the ejection of a neutron alters its composition. Understanding the properties of a compound nucleus is vital for predicting the products of nuclear reactions and for identifying characteristics of nuclear forces and stability within an atom.