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NEET PHYSICSEasy

The mass of a 37Li{}_{3}^{7}\mathrm{Li} nucleus is 0.042 u less than the sum of the masses of its nucleons. The binding energy per nucleon of 37Li{}_{3}^{7}\mathrm{Li} nucleus is nearly:

A

46 MeV

B

5.6 MeV

C

3.9 MeV

D

23 MeV

Step-by-Step Solution

  1. Identify Given Data: The mass defect (Δm\Delta m) is given directly as 0.042 u0.042 \text{ u}. This is the difference between the sum of the masses of the nucleons and the actual mass of the nucleus.
  2. Calculate Total Binding Energy (BE): The binding energy is the energy equivalent of the mass defect. Using the standard conversion factor 1 u≈931.5 MeV1 \text{ u} \approx 931.5 \text{ MeV} : BE=Δm×931.5 MeV/uBE = \Delta m \times 931.5 \text{ MeV/u} BE=0.042×931.5≈39.12 MeVBE = 0.042 \times 931.5 \approx 39.12 \text{ MeV}
  3. Calculate Binding Energy per Nucleon: The nucleus is Lithium-7 (37Li{}_{3}^{7}\mathrm{Li}), which has a mass number A=7A = 7 (total number of nucleons). BEper nucleon=BEABE_{\text{per nucleon}} = \frac{BE}{A} BEper nucleon=39.12 MeV7≈5.58 MeVBE_{\text{per nucleon}} = \frac{39.12 \text{ MeV}}{7} \approx 5.58 \text{ MeV}
  4. Conclusion: The calculated value is approximately 5.6 MeV5.6 \text{ MeV}, which matches Option 2.
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