Problem 3 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

\({ }^{23} \mathrm{Na}\) is the more stable isotope of \(\mathrm{Na}\). Find out the process by which \({ }_{1}^{24}\) Na can undergo radioactive decay (a) \(\beta\) - emission (b) \(\alpha\) emission (c) \(\beta^{+}\)emission (d) \(\mathrm{K}\) electron capture

Step-by-Step Solution

Answer

\(^{24}_{11}\text{Na}\) undergoes \(\beta^-\) emission.

Step 1: Understanding the Problem

We need to determine the possible radioactive decay process for the isotope \({ }^{24}_{11}\text{Na}\). This requires understanding which process is feasible based on its nuclear characteristics.

Step 2: Identifying Characteristics of \(^{24}_{11}\text{Na}\)

\(^{24}_{11}\text{Na}\) is a neutron-rich isotope, meaning it has more neutrons than protons. The isotope has 13 neutrons (24-11) and 11 protons, making it unstable.

Step 3: Possible Decay Processes

Radioactive decay processes include \(\beta^-\) emission, \(\alpha\) emission, \(\beta^+\) emission, and electron capture. \(\beta^-\) emission is most common for neutron-rich isotopes as it allows conversion of neutrons to protons.

Step 4: Analyzing Each Decay Option

For \(\beta^-\) emission, a neutron turns into a proton, emitting an electron and an antineutrino. \(\alpha\) emission generally occurs in heavy elements. \(\beta^+\) emission (positron emission) and electron capture typically occur in proton-rich isotopes.

Step 5: Selecting the Feasible Option

Given \(^{24}_{11}\text{Na}\) has excess neutrons, \(\beta^-\) emission is the likely decay process. This process reduces the neutron count by converting a neutron to a proton, stabilizing the nucleus.

Key Concepts

Beta-minus decay

Beta-minus decay is a form of radioactive decay commonly occurring in neutron-rich isotopes. During this process, a neutron in the nucleus is converted into a proton. This conversion happens because a **W** boson mediates the transformation of a down quark into an up quark, resulting in the emission of a beta particle.

The beta particle emitted is essentially an electron, and alongside it, an antineutrino is also released. This decay results in

• The increase of the atomic number by one unit since a neutron turns into a proton
• The mass number remains the same because the total number of nucleons is unchanged

Beta-minus decay is vital for achieving stability, especially in isotopes like \(^{24}_{11}\text{Na}\), which have an excess of neutrons.

Overall, this decay enhances nuclear stability by converting neutrons to protons, adjusting the neutron-to-proton ratio towards a more balanced state.

Neutron-rich isotopes

Neutron-rich isotopes are atoms whose nuclei have more neutrons compared to protons. This imbalance often leads to instability. Typically, stable isotopes have a well-balanced neutron-to-proton ratio.

In neutron-rich isotopes:

• The excess neutrons may cause the nucleus to be unstable due to increased nuclear forces trying to keep the nucleons together
• Such isotopes may undergo decay processes, like beta-minus decay, to achieve a more stable form

For instance, \(^{24}_{11}\text{Na}\) is neutron-rich with 13 neutrons and 11 protons, making it unstable. This imbalance prompts the isotope to undergo beta-minus decay, converting a neutron into a proton.

By doing so, the isotope emerges more balanced and stable by reducing the neutron count. Balancing the forces within the nucleus is crucial as it directly relates to the concept of nuclear stability.

Nuclear stability

Nuclear stability refers to how likely an atomic nucleus is to remain unchanged or "stable" over time. Several factors influence nuclear stability, including the neutron-to-proton ratio, the presence of magic numbers, and the overall binding energy.

Isotopes with a favorable neutron-proton ratio are often more stable. However, if there are too many neutrons or protons, the nucleus can become unstable, leading to radioactive decay. Here's what makes a nucleus stable or unstable:

• The neutron number must be enough to buffer repulsive forces among protons.
• Magic numbers of nucleons are extra-stable configurations where the binding energies are higher.

For neutron-rich isotopes like \(^{24}_{11}\text{Na}\), balancing this ratio via processes such as beta-minus decay helps bring about nuclear stability.

In essence, isotopes strive for a stable configuration where the energy inside the nucleus represents the most stable state possible. This explains why isotope transformation is crucial: it's a transformation towards greater stability, which is the ultimate goal of radioactive decay processes.