Question

Consider the following decay \({ }_{z} X^{A} \rightarrow_{z+1} Y^{A}+{ }_{-1} e^{0}, X\) is unstable because : (a) its nucleus has excess energy (b) \(\frac{n}{p}\) ratio is high (c) \(\frac{n}{p}\) ratio is low (d) none of these

Short answer

(b) the n/p ratio is high.

Step-by-step solution

Identify the Type of Decay

Examine the given nuclear equation to identify the type of decay. Here, \({ }_{z} X^{A} \rightarrow_{z+1} Y^{A}+{ }_{-1} e^{0}\) represents beta-minus decay (β− decay), where a neutron in the nucleus of an atom decays into a proton, emitting an electron (beta particle) in the process.

Understand Beta-minus Decay

Understand that in beta-minus decay, a neutron is converted to a proton, electron, and antineutrino. This type of decay typically occurs in nuclei with a high neutron-to-proton (n/p) ratio, as it allows the nucleus to move towards a more stable configuration with a higher proton count.

Determine the Correct Reason for Instability

Given the type of decay, the instability of the initial nucleus (X) can be attributed to a high neutron-to-proton ratio. Since beta-minus decay occurs to reduce the number of neutrons, it is the high n/p ratio that causes the nucleus to be unstable.

Select the Correct Option

Based on the understanding of beta-minus decay, the correct option that explains the instability of the nucleus X is (b) the n/p ratio is high.

Key concepts

Nuclear Decay

Nuclear decay is the process by which an unstable atomic nucleus loses energy by emitting radiation. There are several types of nuclear decay, and each type involves the release of different particles or electromagnetic waves. Beta-minus decay is a common form of radioactive decay where an unstable nucleus emits a beta particle (an electron) alongside an antineutrino.

During beta-minus decay, a neutron in the nucleus is transformed into a proton. This increases the atomic number of the element by one, resulting in the formation of a different element. For instance, a carbon-14 nucleus ({^{14}_{6}C}) undergoing beta-minus decay will become a nitrogen-14 nucleus ({^{14}_{7}N}), effectively changing the identity of the element involved.

One of the most fascinating aspects of nuclear decay is its role in various natural phenomena. For example, beta-minus decay is central to the process of carbon dating, which archaeologists use to determine the age of ancient artifacts. Moreover, this type of decay has a crucial function in the field of medicine, specifically in nuclear imaging techniques such as Positron Emission Tomography (PET).

Neutron-to-Proton Ratio

The neutron-to-proton ratio ({n/p}) within an atomic nucleus is a crucial determinant of nuclear stability. Atoms strive for a balance between the number of protons and neutrons to achieve the most stable configuration. Too many or too few neutrons in relation to protons can make an atom unstable, often leading to nuclear decay.

In lighter elements (up to iron), the most stable nuclei typically have a nearly equal number of protons and neutrons. As the elements get heavier, a higher neutron-to-proton ratio is needed for stability. This is due to the increasing electrostatic repulsion between protons that must be counterbalanced by a larger number of neutrons.

If the {n/p} ratio is too high, meaning there are too many neutrons, the nucleus may undergo beta-minus decay to convert a neutron into a proton and achieve a more favorable balance. On the other hand, if the {n/p} ratio is too low (too few neutrons), the nucleus may undergo beta-plus decay or electron capture to increase the number of neutrons, although this is not the case in the given exercise.

Nuclear Stability

Nuclear stability is driven by the delicate interplay between the strong nuclear force that binds protons and neutrons together, and the disruptive electrostatic forces among protons in the nucleus. Stable nuclei are characterized by an optimal balance in the strong nuclear force and electrostatic repulsion, usually achieved through a favorable neutron-to-proton ratio.

Most stable isotopes reside within the 'Belt of Stability' on the chart of nuclides, which graphs neutron number against proton number. Atoms outside this belt are inherently unstable and will tend to transform via nuclear decay into a more stable configuration within the belt.

The beta-minus decay addressed in the exercise helps to stabilize an unstable nucleus with a high neutron-to-proton ratio, moving it towards the Belt of Stability. It is one of nature's ways of correcting an imbalance and achieving nuclear stability. Hence, when an atom like {X} in our exercise exhibits beta-minus decay, it is a direct response to the need to attain a more energetically favorable state—with a improved balance between neutrons and protons.