Question 5 of 13intermediate🔧 ApplyNumerical2 marks

Obtain the binding energy (in MeV) of a nitrogen nucleus 714N^{14}_{7}\text{N}, given m(714N)=14.00307um(^{14}_{7}\text{N}) = 14.00307 \text{u}

Correct Answer

The binding energy of the nitrogen nucleus 714N^{14}_{7}\text{N} is 104.67 MeV.

Exercise: EXERCISES | Q: 13.1 | (Page 16)
For More Understanding

Explanation

This question tests the concept of nuclear binding energy calculation using mass defect. The binding energy is the energy equivalent of the mass defect, which is the difference between the total mass of individual nucleons and the actual mass of the nucleus. The conversion factor 1 u = 931.5 MeV is provided in the textbook context.

Solution Steps

  1. Step 1: Identify the number of protons and neutrons in 714N^{14}_{7}\text{N}. Atomic number Z = 7, Mass number A = 14. Number of neutrons = A - Z = 14 - 7 = 7.

  2. Step 2: Calculate the total mass of constituent nucleons. Mass of 7 protons = 7 × mHm_H = 7×1.0078257 \times 1.007825 = 7.054775 u. Mass of 7 neutrons = 7 × mnm_n = 7×1.0086657 \times 1.008665 = 7.060655 u. Total mass of nucleons = 7.054775 + 7.060655 = 14.11543 u.

  3. Step 3: Calculate the mass defect (ΔM\Delta M). ΔM\Delta M = Total mass of nucleons - Mass of nucleus = 14.11543 - 14.00307 = 0.11236 u.

  4. Step 4: Convert mass defect to binding energy using the conversion factor. Binding Energy = ΔM\Delta M × 931.5 MeV/u = 0.11236×931.50.11236 \times 931.5 = 104.67 MeV.

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