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The nucleus 6C12 consists of six protons and six neutrons held in close association by strong nuclear forces
The nucleus 6C12 consists of six protons and six neutrons held in close association by strong nuclear forces. The atomic rest masses are
6C12: 12.000000u
1H1: 1.007825u
n: 1.008665u
in terms of the atomic mass unit u = 1.66 x 10^-27 kg. How much energy would be required to separate a 6C12 nucleus into its constituent protons and neutrons?
This problem has me totally confused. Am I using the formula for Relativistic kinetic energy:
K = (mc2/√1 - v2/c2) - mc2
or for rest energy:
E0 = mc2
or neither? How do I do this problem?
Expert Solution
The nucleus 6C12 consists of six protons and six neutrons held in close association by strong nuclear forces. The atomic rest masses are
6C12: 12.000000u
1H1: 1.007825u
n: 1.008665u
in terms of the atomic mass unit u = 1.66 x 10-27 kg. How much energy would be required to separate a 6C12 nucleus into its constituent protons and neutrons?
Mass of 1 proton = 1.007825u
Therefore, mass of 6 protons = 6*1.007825u = 6.04695u
Mass of 1 neutron = 1.008665u
Therefore, mass of 6 neutrons = 6*1.008665u = 6.05199u
Total mass = 6.04695u + 6.05199u = 12.09894u
Therefore, ?m = 12.09894u - 12.000000u = 0.09894u
But 1 u = 931.494 MeV/c2
Nuclear binding energy = ?mc2 = 0.09894*931.494 MeV = 92.162016 MeV
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