Problem 90 - A Complete Resource Book in Chemistry for JEE Main

Question

\({ }_{13} \mathrm{Al}^{27}\) is a stable isotope. \({ }_{13} \mathrm{Al}^{29}\) is expected to disintegrate by (a) \(\alpha\) emission (b) \(\beta\) emission (c) positron emission (d) proton emission

Step-by-Step Solution

Answer

\({ }_{13} \mathrm{Al}^{29} \) is likely to disintegrate by \(\beta\) emission (b).

Step 1: Identify the Type of Isotopes

First, identify the nature of both isotopes presented in the problem. Both belong to Aluminum (Al), which has an atomic number of 13.

Step 2: Calculate Neutron Counts

Calculate the number of neutrons in both isotopes. For \({ }_{13} \mathrm{Al}^{27}\), the number of neutrons is \(27 - 13 = 14\). For \({ }_{13} \mathrm{Al}^{29}\), the number of neutrons is \(29 - 13 = 16\).

Step 3: Evaluate Stability: Neutron-Proton Ratio

Compare the neutron-to-proton ratio for both isotopes. In general, increased neutron numbers compared to protons suggest beta decay if the isotope is unstable. For Aluminum-29, there are 16 neutrons and 13 protons, suggesting possible instability due to excess neutrons.

Step 4: Determine Likely Decay Mode

Given an excess of neutrons, Aluminum-29 would most likely undergo \(\beta\) decay to convert a neutron into a proton, emitting an electron (beta particle) and an antineutrino. This process reduces the neutron-to-proton ratio, moving the nucleus toward a more stable state.

Key Concepts

Isotopes

Isotopes are variants of the same chemical element containing the same number of protons but different numbers of neutrons in their nuclei. This difference in neutrons can lead to variations in the mass of the isotopes, as the atomic number remains constant across isotopes of a given element. For example, aluminum (\( \mathrm{Al} \)) has an atomic number of 13, meaning each of its isotopes has 13 protons.
A stable isotope means it has a balanced configuration of neutrons and protons that do not react spontaneously over time. For instance, \({ }_{13} \mathrm{Al}^{27}\), an isotope of aluminum, is stable because it maintains such a balanced configuration. On the other hand, \({ }_{13} \mathrm{Al}^{29}\) is an unstable isotope due to an imbalance of neutrons and protons, leading to potential radioactive decay.

• Stable isotopes do not easily undergo radioactive decay.
• Unstable isotopes may emit radiation to reach stability.
• The difference between isotopes is crucial for determining their potential decay modes.

Neutron-Proton Ratio

The neutron-proton ratio is a critical factor in determining the stability of an isotope. This ratio indicates the balance between the number of neutrons and protons within an atomic nucleus. For many elements, particularly lighter ones, a 1:1 ratio signifies stability. However, as the atomic number increases, a greater number of neutrons is generally needed to stabilize the nucleus.
In our specific example of aluminum isotopes:

• \({ }_{13} \mathrm{Al}^{27}\), which is stable, has a neutron-proton ratio of approximately 1.08 (14 neutrons to 13 protons).
• \({ }_{13} \mathrm{Al}^{29}\) has a higher ratio with 16 neutrons to 13 protons, equaling about 1.23. This higher ratio suggests instability, often leading to beta decay.

Understanding neutron-proton ratios helps predict which isotopes are likely unstable and therefore keen on undergoing certain decay processes to achieve balance, such as beta decay.

Beta Decay

Beta decay is a type of radioactive decay involving the emission of beta particles from an unstable nucleus. It occurs in isotopes with a neutron-proton ratio that is too high, causing instability. During beta decay, a neutron transforms into a proton, an electron (beta particle), and an antineutrino. The conversion effectively lowers the neutron-proton ratio, helping the nucleus attain a more stable arrangement.
Here's how beta decay operates:

• An excess neutron in the nucleus is converted into a proton.
• The process emits an electron and an antineutrino, reducing the neutron count by one and increasing the proton count by one.
• This shift balances the neutron-proton ratio, moving the isotope toward stability.

In our example, \({ }_{13} \mathrm{Al}^{29}\) experiences beta decay due to its imbalanced neutron-proton ratio. Through beta decay, it transitions towards a stable form, potentially avoiding further spontaneous radioactive behavior.