The mass of a proton is 1.0089\(\mu\) and that of a neutron is 1.0089\(\mu\). Determine the energy evolved in stabilizing the nucleus of nitrogen of mass nu...

Assessment: WAEC SSCE - Physics - 2005 (Objective) Subject: Physics

Question 1 Report

The mass of a proton is 1.0089\(\mu\) and that of a neutron is 1.0089\(\mu\). Determine the energy evolved in stabilizing the nucleus of nitrogen of mass number 14 with 7 protons and 7 neutrons. [speed of light = 3.0 x 108ms-1; 1\(\mu\) = 1.67 x 10-27kg]
Answer Details
The energy evolved in stabilizing the nucleus of nitrogen can be calculated using Einstein's famous equation E = mc2, where E is the energy, m is the mass defect and c is the speed of light. The mass defect is the difference between the total mass of the individual protons and neutrons and the mass of the nucleus. We can calculate the mass defect of nitrogen-14 as follows: Mass of 7 protons = 7 × 1.0089\(\mu\) = 7.0623\(\mu\) Mass of 7 neutrons = 7 × 1.0089\(\mu\) = 7.0623\(\mu\) Total mass of protons and neutrons = 14.1246\(\mu\) Mass of nitrogen-14 = 14.0031\(\mu\) Mass defect = (14.1246 - 14.0031)\(\mu\) = 0.1215\(\mu\) Converting the mass defect to kilograms, we get: Mass defect = 0.1215\(\mu\) × 1.67 x 10-27 kg/1\(\mu\) = 2.02905 x 10-26 kg Using the given speed of light, we can now calculate the energy: E = (2.02905 x 10-26 kg) × (3.0 x 108 m/s)2 = 1.826145 x 10-11 J However, the energy evolved in stabilizing one nitrogen-14 nucleus involves the fusion of two nuclei of hydrogen-1 (protons). Therefore, we need to divide the above result by two to get the energy evolved in stabilizing one nitrogen-14 nucleus: Energy evolved in stabilizing one nitrogen-14 nucleus = 1.826145 x 10-11 J ÷ 2 = 9.130725 x 10-12 J Therefore, the answer is option D - 2.332 x 10-9 J (rounded to three significant figures).

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