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Staements I: Light nuclei are most stabl...

Staements I: Light nuclei are most stable if N =Z, while heavy nuclei are more stable if N `gt` Z. (N`rarr` number of neutrons, Z`rarr` number of protons)
Staements II: As the number of protons increases in a nucleus, the Coulomb's repulsive forece increases, which tends to break the nucleus apart. So, to keep the nucleus apart. So, to keep the nucleus stable, more number of neutrons are needed which are neutral in nature.

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To analyze the statements provided in the question, we will break down the reasoning step by step. ### Step 1: Understand Statement I **Statement I** claims that light nuclei are most stable when the number of neutrons (N) equals the number of protons (Z), while heavy nuclei are more stable when the number of neutrons is greater than the number of protons. - **Light Nuclei Stability**: For light nuclei, such as helium (with 2 protons and 2 neutrons), the stability is achieved when N = Z. This is because the strong nuclear force, which binds protons and neutrons together, is sufficient to overcome the repulsive Coulomb force between protons when they are in equal numbers. - **Heavy Nuclei Stability**: As we move to heavier nuclei, the number of protons increases, leading to a stronger Coulomb repulsion. To counterbalance this repulsion and maintain stability, more neutrons are required (N > Z). Neutrons help to add nuclear binding without contributing to the repulsive force. **Conclusion for Statement I**: True. ### Step 2: Understand Statement II **Statement II** states that as the number of protons increases in a nucleus, the Coulomb repulsive force increases, which tends to break the nucleus apart. To keep the nucleus stable, more neutrons are needed since they are neutral. - **Coulomb Repulsion**: As the number of protons increases, the Coulomb repulsive force indeed increases. This repulsion acts between the positively charged protons, making it harder for the nucleus to remain stable. - **Role of Neutrons**: Neutrons do not carry any charge, so they do not contribute to the Coulomb repulsion. Instead, they contribute to the strong nuclear force, which helps to bind the nucleus together. Thus, to maintain stability in heavier nuclei, an increase in the number of neutrons is necessary to offset the increasing repulsive forces among protons. **Conclusion for Statement II**: True, and it correctly explains the reasoning behind Statement I. ### Final Conclusion Both statements are true, and Statement II provides the correct explanation for Statement I.

To analyze the statements provided in the question, we will break down the reasoning step by step. ### Step 1: Understand Statement I **Statement I** claims that light nuclei are most stable when the number of neutrons (N) equals the number of protons (Z), while heavy nuclei are more stable when the number of neutrons is greater than the number of protons. - **Light Nuclei Stability**: For light nuclei, such as helium (with 2 protons and 2 neutrons), the stability is achieved when N = Z. This is because the strong nuclear force, which binds protons and neutrons together, is sufficient to overcome the repulsive Coulomb force between protons when they are in equal numbers. - **Heavy Nuclei Stability**: As we move to heavier nuclei, the number of protons increases, leading to a stronger Coulomb repulsion. To counterbalance this repulsion and maintain stability, more neutrons are required (N > Z). Neutrons help to add nuclear binding without contributing to the repulsive force. ...
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Statement I:Heavy nuclides tend to have more number of neutrons than protons. Staements II: In hevay nuclei, as there is coloumbic repulsion between protons, so excess of neutrons are preferable: