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Question: The binding energies of the nuclei of $^4_2He$, $^7_3Li$, $^{12}_6C$ & $^{14}_7N$ are 28, 52, 90, 98...

The binding energies of the nuclei of 24He^4_2He, 37Li^7_3Li, 612C^{12}_6C & 714N^{14}_7N are 28, 52, 90, 98 MeV respectively.

Which of these is most stable.

A

24He^4_2He

B

37Li^7_3Li

C

612C^{12}_6C

D

714N^{14}_7N

Answer

The most stable nucleus is 612C^{12}_6C.

Explanation

Solution

To determine the most stable nucleus, we need to calculate the binding energy per nucleon (BE/A) for each given nucleus. A higher binding energy per nucleon indicates greater stability.

The number of nucleons (A) for each nucleus is given by the superscript.

  1. For 24He^4_2He:

    • Binding Energy (BE) = 28 MeV
    • Number of nucleons (A) = 4
    • Binding Energy per Nucleon = 28 MeV4 nucleons=7.0 MeV/nucleon\frac{28 \text{ MeV}}{4 \text{ nucleons}} = 7.0 \text{ MeV/nucleon}

  2. For 37Li^7_3Li:

    • Binding Energy (BE) = 52 MeV
    • Number of nucleons (A) = 7
    • Binding Energy per Nucleon = 52 MeV7 nucleons≈7.43 MeV/nucleon\frac{52 \text{ MeV}}{7 \text{ nucleons}} \approx 7.43 \text{ MeV/nucleon}

  3. For 612C^{12}_6C:

    • Binding Energy (BE) = 90 MeV
    • Number of nucleons (A) = 12
    • Binding Energy per Nucleon = 90 MeV12 nucleons=7.5 MeV/nucleon\frac{90 \text{ MeV}}{12 \text{ nucleons}} = 7.5 \text{ MeV/nucleon}

  4. For 714N^{14}_7N:

    • Binding Energy (BE) = 98 MeV
    • Number of nucleons (A) = 14
    • Binding Energy per Nucleon = 98 MeV14 nucleons=7.0 MeV/nucleon\frac{98 \text{ MeV}}{14 \text{ nucleons}} = 7.0 \text{ MeV/nucleon}

Comparing the binding energies per nucleon:

  • 24He^4_2He: 7.0 MeV/nucleon
  • 37Li^7_3Li: 7.43 MeV/nucleon
  • 612C^{12}_6C: 7.5 MeV/nucleon
  • 714N^{14}_7N: 7.0 MeV/nucleon

The nucleus 612C^{12}_6C has the highest binding energy per nucleon (7.5 MeV/nucleon), making it the most stable among the given options.