Nucleons The term nucleon refers to the particles found in the nucleus of an ato

Question

Nucleons The term nucleon refers to the particles found in the nucleus of an atom, namely protons and neutrons. A single hydrogen atom (one proton plus one electron) has a mass of 1.007825 amu. A single neutron has a mass of 1.008665 amu. Note that amu stands for "atomic mass unit" and is sometimes abbreviated with the symbol u. 1 u is equivalent to 1.6605387×1027 kg. Mass Defect The total mass of the individual nucleons in an atom is always greater than the actual measured mass of the atom. The difference is called the mass defect. Binding Energy The mass defect of an atom corresponds to an amount of energy called the binding energy. Binding energy can be calculated using Einstein's equation, E=mc2., where E is energy in joules, m is mass is kilograms, and c is the speed of light, 2.998×108 m/s.

Part A: Calculate the mass defect of the nitrogen nucleus 14 7N. The mass of neutral 14 7N is equal to 14.003074 atomic mass units.

answer: .1124 amu

Part B: Calculate the binding energy E of the nitrogen nucleus 14 7N (1eV=1.602×1019J). answer: 104.7 MeV Part C: Calculate the binding energy per nucleon of the nitrogen nucleus 14 7N.

answer: 7.477 MeV/nucleon

Part D: Calculate the mass defect of the helium nucleus 52He. The mass of neutral 52He is given by MHe=5.012225amu.

Part E: Calculate the binding energy E of the helium nucleus 52He (1eV=1.602×1019J).

Part F: Calculate the binding energy E of the helium nucleus 52He (1eV=1.602×1019J).

Explanation / Answer

A) The mass of 14N7 is 14.003074 u. 14N7 has 14 nucleons and 7 protons; hence there are (14 – 7) = 7 neutrons.

The sum of the masses of the protons and neutrons in 14N7 is (7*1.007825 + 7*1.008665) u = 14.11543 u (mass of proton = mass of hydrogen atom containing one proton and electron; mass of an electron is considered negligible).

The mass defect in 14N7 = (14.11543 – 14.003074) u = 0.112356 u 0.1124 u (ans).

B) The binding energy E of the nucleus in 14N7 is E = m*c2 = (0.1124 u)*(2.998*108 m/s)2 = (0.1124 u)*(1.6605387*10-27 kg/1 u)*(2.998*108 m/s)2 = 1.677562*10-11 kg.m2.s-2 = (1.677562*10-11 kg.m2.s-2)*(1 J/1 kg.m2.s-2)*(1 eV/1.602*10-19 J) = 1.04716*108 eV = (1.04716*108 eV)*(1 MeV/1.0*106 eV) = 104.716 MeV 104.7 MeV (ans).

C) There are 14 nucleons; hence the binding energy per nucleon E is (104.7 MeV)/(14) = 7.47857 MeV 7.477 MeV (ans).

D) The mass of 5He2 is 5.012225 u. 5He2 has 5 nucleons and 2 protons; hence there are (5 – 7) = 3 neutrons.

The sum of the masses of the protons and neutrons in 5He2 is (2*1.007825 + 3*1.008665) u = 5.041645 u.

The mass defect in 5He2 = (5.041645 – 5.012225) u = 0.02942 u (ans).

E) The binding energy E of the nucleus in 5He2 is E = m*c2 = (0.02942 u)*(2.998*108 m/s)2 = (0.02942 u)*(1.6605387*10-27 kg/1 u)*(2.998*108 m/s)2 = 4.390914*10-12 kg.m2.s-2 = (4.390914*10-12 kg.m2.s-2)*(1 J/1 kg.m2.s-2)*(1 eV/1.602*10-19 J)*(1 MeV/1.0*106 eV) = 27.40895 MeV 27.41 MeV (ans).

F) There are 5 nucleons; hence the binding energy per nucleon E is (27.41 MeV)/(5) = 5.482 MeV 5.48 MeV (ans).

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