Statement 1: The nuclide _{13}^{30}Al is less | vedclass

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(A) For $_{13}^{30}Al$: Number of protons $(Z)$ = $13$ (odd),Number of neutrons $(N)$ = $30 - 13 = 17$ (odd). For $_{20}^{40}Ca$: Number of protons $(

Vedclass · Accepted Answer

Statement $1$ and Statement $2$ are both true and Statement $2$ is the correct explanation of Statement $1$.

Vedclass · Answer

(A) For $_{13}^{30}Al$: Number of protons $(Z)$ = $13$ (odd),Number of neutrons $(N)$ = $30 - 13 = 17$ (odd). For $_{20}^{40}Ca$: Number of protons $(Z)$ = $20$ (even),Number of neutrons $(N)$ = $40 - 20 = 20$ (even). Nuclides with even numbers of both protons and neutrons are generally more stable than those with odd numbers of both. Since $_{13}^{30}Al$ has odd-odd nucleons and $_{20}^{40}Ca$ has even-even nucleons,$_{13}^{30}Al$ is less stable. Statement $2$ is a general rule regarding nuclear stability based on the odd-even nature of nucleons.

Statement $1$: The nuclide $_{13}^{30}Al$ is less stable than $_{20}^{40}Ca$. Statement $2$: Nuclides containing an odd number of neutrons are generally unstable.

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