Problem 120 - A Complete Resource Book in Chemistry for JEE Main

Question

Which of the following pairs are isodiapheric pairs? (a) \({ }_{29} \mathrm{Cu}^{65}\) and \({ }_{24} \mathrm{Cr}^{55}\) (b) \({ }_{29} \mathrm{Cu}^{65}\) and \({ }_{24} \mathrm{Cr}^{52}\) (c) \({ }_{92}^{2} \mathrm{U}^{235}\) and \({ }_{90}^{24} \mathrm{Th}^{231}\) (d) \({ }_{92} \mathrm{U}^{238}\) and \({ }_{90}^{24} \mathrm{Th}^{231}\)

Step-by-Step Solution

Answer

Pairs (a) and (c) are isodiapheric.

Step 1: Understanding Isodiapheric Pairs

Isodiapheric pairs are nuclei pairs that have the same difference between the number of neutrons and protons. To identify isodiapheric pairs, we calculate the difference \( N - Z \), where \( N \) is the number of neutrons and \( Z \) is the number of protons.

Step 2: Calculate Neutron-Proton Difference for Cu-65 and Cr-55

For \( \mathrm{Cu}_{29}^{65} \), neutrons \( N_1 = 65 - 29 = 36 \). Hence, \( N_1 - Z_1 = 36 - 29 = 7 \).For \( \mathrm{Cr}_{24}^{55} \), neutrons \( N_2 = 55 - 24 = 31 \). Hence, \( N_2 - Z_2 = 31 - 24 = 7 \).Thus, \( \mathrm{Cu}_{29}^{65} \) and \( \mathrm{Cr}_{24}^{55} \) are isodiapheric pairs.

Step 3: Calculate Neutron-Proton Difference for Cu-65 and Cr-52

For \( \mathrm{Cu}_{29}^{65} \), we have already calculated that \( N_1 - Z_1 = 7 \).For \( \mathrm{Cr}_{24}^{52} \), neutrons \( N_2 = 52 - 24 = 28 \). Hence, \( N_2 - Z_2 = 28 - 24 = 4 \).Thus, \( \mathrm{Cu}_{29}^{65} \) and \( \mathrm{Cr}_{24}^{52} \) are not isodiapheric pairs.

Step 4: Calculate Neutron-Proton Difference for U-235 and Th-231

For \( \mathrm{U}_{92}^{235} \), neutrons \( N_1 = 235 - 92 = 143 \). Hence, \( N_1 - Z_1 = 143 - 92 = 51 \).For \( \mathrm{Th}_{90}^{231} \), neutrons \( N_2 = 231 - 90 = 141 \). Hence, \( N_2 - Z_2 = 141 - 90 = 51 \).Thus, \( \mathrm{U}_{92}^{235} \) and \( \mathrm{Th}_{90}^{231} \) are isodiapheric pairs.

Step 5: Calculate Neutron-Proton Difference for U-238 and Th-231

For \( \mathrm{U}_{92}^{238} \), neutrons \( N_1 = 238 - 92 = 146 \). Hence, \( N_1 - Z_1 = 146 - 92 = 54 \).For \( \mathrm{Th}_{90}^{231} \), we have already calculated that \( N_2 - Z_2 = 51 \).Thus, \( \mathrm{U}_{92}^{238} \) and \( \mathrm{Th}_{90}^{231} \) are not isodiapheric pairs.

Key Concepts

Neutron-proton difference

In the world of nuclear chemistry, an important concept is the difference between the number of neutrons and protons in an atomic nucleus. This difference is crucial when identifying isodiapheric pairs. An isodiapheric pair of nuclei means that both nuclei exhibit the same neutron-proton difference. To find this difference, we subtract the number of protons, denoted by \( Z \), from the number of neutrons, \( N \). The calculation looks like this: \( N - Z \).
Calculating this quantity helps us understand whether two different isotopes behave similarly when participating in nuclear reactions. This understanding is particularly useful in predicting nuclear stability and potential energy changes that occur during such reactions.
Identifying isodiapheric pairs means performing this operation on different isotopes, as shown in the original exercise, where each pair's neutron-proton difference was calculated to determine if they shared the same value.

Nuclear chemistry

Nuclear chemistry is a branch of chemistry that deals with the nuclei of atoms. Unlike regular chemistry, which focuses on electron interactions, nuclear chemistry primarily looks at how the protons and neutrons within the nucleus interact with one another.
It encompasses everything from the behavior of radioactive elements, nuclear fission, and fusion, to the roles isotopes play. Understanding isodiapheric pairs is a small, but significant part of nuclear chemistry. It gives insights into nuclear reactions stability and the force balance within a nucleus between these subatomic particles.
In particular, nuclear reactions are sensitive to the balance of neutrons and protons in a nucleus. Therefore, calculating differences, as practiced in determining isodiapheric pairs, can help predict how atoms will react during nuclear changes such as decay or when a nucleus is bombarded in an accelerator.

Atomic structure

The atomic structure is the foundation of all nuclear chemistry discussions. An atom comprises a central nucleus made up of protons and neutrons, surrounded by electrons in various energy levels. The number of protons in an atom's nucleus, referred to as the atomic number, defines what element the atom belongs to.
Neutrons add mass to the nucleus without affecting the chemical properties, but they do play a crucial role in the stability and type of isotope. Different isotopes of an element have the same number of protons but differ in the number of neutrons.
Understanding the balance of protons and neutrons helps in many aspects such as determining isodiapheric pairs. Knowing how many neutrons an isotope has compared to its protons aids in predicting its behavior in nuclear processes, stability, and even potential uses in medical or energy-producing applications.

Neutron calculation

Calculating the number of neutrons in an atom is fundamental in many areas of chemistry and physics. To calculate the number of neutrons in an atom, we subtract the atomic number \( Z \) (the number of protons) from the atomic mass number (the total amount of protons and neutrons) of the isotope.
For example, to determine the number of neutrons in a copper-65 isotope, labeled as \( _{29}^{65} \text{Cu} \), subtract 29 from 65, giving us 36 neutrons. This type of calculation is essential when identifying isodiapheric pairs.
Ensuring accuracy in these calculations is vital in predicting nuclear reactions and understanding more about the elemental properties. This information does not only serve academic purposes but also aids in real-world applications, including the safe handling of nuclear materials and the innovation of nuclear energy systems.