Which of the following is/are correct statements? — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - Evaluate Statement (A) "All nuclei having n/p ratio between 1.0 and 1.52 have stable nuclei." This is **incorrect**. Many radioactive nuclides have n/p ratios in this range. For example, Carbon-14 has n/p = 8/6 = 1.33, but it is radioactive with $t_{1/2} = 5730$ yr. Stability depends on multiple factors beyond just the n/p ratio. Step 2 - Evaluate Statement (B) "All nuclei having binding energy per nucleon greater than that of $\ce{^{209}_{83}Bi}$ have stable nuclei." $\ce{^{209}Bi}$ is the heaviest stable nuclide. Nuclei heavier than Bi (with $Z > 83$) are all radioactive. However, the binding energy per nucleon actually peaks around Fe-56 ($\sim$8.8 MeV) and decreases for very heavy nuclei. No nuclide heavier than Bi-209 has a higher binding energy per nucleon, so the statement is effectively correct: all stable nuclei (except Bi-209 itself) have intermediate binding energies. → **Correct.** ✓ Step 3 - Evaluate Statement (C) "All nuclei having even number of protons as well as neutrons (even-even) have stable nuclei." This is **incorrect**. While even-even nuclei tend to be more stable, several even-even nuclides are radioactive (e.g., Polonium-210 is alpha-radioactive despite being even-Z). The statement is too absolute. Step 4 - Evaluate Statement (D) "There are very few stable nuclei having odd number of protons as well as neutrons (odd-odd)." This is **correct**. Only 4 stable odd-odd nuclei exist: $\ce{^2_1H}$, $\ce{^6_3Li}$, $\ce{^{10}_5B}$, and $\ce{^{14}_7N}$. All other odd-odd nuclides are radioactive. → **Correct.** ✓ $$\boxed{\text{Answer: B, D}}$$