Protons, Neutrons and Electrons

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Atom P’s, N’s and e’s oh please!! Atomic Number (Z) It is the smaller, whole number in a square on the periodic table for one element. It states the number of protons. Mass number (A) The mass number is the larger number with a decimal. The number states the total number of protons and neutrons the average atom of that element has. The definition of an element is determined by the number of protons. To determine the number of neutrons, you need to do a little subtraction. First round the mass number to the nearest whole number (because you either have a proton or neutron or you do not) Mass Number (A) – Atomic Number (Z) = neutrons Practice! How many protons and neutrons does Ca (calcium) have? Mass number is 40 Atomic number is 20 40-20 = 20 Practice Individually Finding Protons and Neutrons Mg Mn • Mg • 24-12 = 12 • Protons are 12 (because of the atomic number) • Neutrons are 12 (because of the subtraction) Mn Atomic number is 25 Mass number rounded is 55 25 protons 55-25 = 30 neutrons Electrons To determine the number of electrons, you need to know if you have a charged atom (it is an ion) or a neutron atom. If it is neutral, then protons = electrons. Neutral Ca 20 protons from its atomic number (Z) Therefore it has 20 electrons Ions An ion is a neutral atom which has gained or lost an electron. An ion with a negative charge is called an anion. (it has extra electrons) An ion with a positive charge is called a cation. So Ca (Calcium) looses two electrons, what will be the resulting charge? +2 Ca has 20 protons, so neutral it would have 20 electrons, but according to the charge, 2 electrons have been lost. Therefore, it will have 20-2 = 18 electrons What if N (Nitrogen) gained three electrons? Its resulting charge is: -3 Nitrogen has 7 electrons when it is neutral. It will need three more electrons to have a -3 charge. 7+3 = 10 Element Protons Neutrons Electrons Charge Ar 15 Mo As 11 0 O -2 Write the symbol When you write the atomic symbol you do the following .

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Mass Defect & Binding Energy
Sem-3 CC7 Nuclear Physics Mass Defect & Binding Energy According to nuclear particle experiments, the total mass of a nucleus (mnuc) is less than the sum of the masses of its constituent nucleons (protons and neutrons). For a nucleus with Z protons and mass number A, the mass difference or mass defect is given by Δm= Zmp + (A−Z)mn − mnuc where Zmp is the total mass of the protons, (A−Z)mn is the total mass of the neutrons, and mnuc is the mass of the nucleus. According to Einstein’s special theory of relativity, mass is a measure of the total energy of a system (E=mc2). The energy equivalent to mass defect is alled the binding energy B of the nucleus. Thus 2 B= [Zmp + (A−Z)mn − mnuc] c The binding energy is equal to the amount of energy released in forming the nucleus. Example 1. Calculate the mass defect and the binding energy of the deuteron. The mass of the −27 deuteron is mD=3.34359×10 kg or 2.014102u. Solution For the deuteron Z=1 and A=2. The mass defect for the deuteron is Δm=mp+mn−mD=1.008665 u+ 1.007825 u- 2.014102u = 0.002388u The binding energy of the deuteron is then B= Δm c2= Δm ×931.5 MeV/u=2.224MeV. Over two million electron volts are needed to break apart a deuteron into a proton and a neutron. This very large value indicates the great strength of the nuclear force. By comparison, the greatest amount of energy required to liberate an electron bound to a hydrogen atom by an attractive Coulomb force (an electromagnetic force) is about 10 eV.
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Iron-57 of Its Isotopes and Has an Sum of Protons + Neutrons 26 Fe Average Mass As Determined by a in the Nucleus
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The Nucleus and Nuclear Instability
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The Delimiting/Frontier Lines of the Constituents of Matter∗
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CHAPTER 2 the Nucleus and Radioactive Decay
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Study of Isotope Shifts, Isotone Shifts and Nuclear Compressibility from the Analysis of Muonic X-Ray Transitions
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Nuclear Physics Sections 10.1-10.7
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Lawrence Berkeley National Laboratory Recent Work
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Nuclear Chemistry
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