Atomic mass unit (amu) and electron volt (eV) In the study of atomic and nuclear physics masses are expressed in atomic mass unit abbreviated amu and energies are expressed in electron volt abbreviated eV. The atomic mass unit is defined as one-twelveth of the mass of the carbon atom gC12. It is expressed in kg (SI unit of mass) as follows : The number of atoms of gC12 in one mole of carbon is 6.0221367 x 1023 (Avagadro number). This means that mass of 6.0 2 2 1 36 7 x 1023 atoms of 6C12 = 12 g = 12 x 10-3 kg. r i ♦ f ^12 12 x 10~3 Mass of 1 atom of 6C =-kg 6.0221367 x10 1 12 1 amu = — x mass of 1 atom of 6C 12 -3 = —x— 12 x 10-— = 1.6605402 x 10~27 kg 12 6.0221367 x 10 Thus, 1 amu (or simply 1 u) is nearly 1.661 x 10~27 kg. The electron volt is defined as the energy acquired by an electron when it is accelerated through a potential difference of 1 volt. It is expressed in joule (SI unit of energy) as follows : 1 electron-volt = 1 eV = (electron charge) (l volt of p. d.) = ^1.60217738 x 10~19 coulomb j (l volt) = 1.60217738 x lO-19 joule Thus, one electron-volt is nearly equal to 1.602 x 10~19 J Relation between amu and eV (1 amu = 931 MeV) Atomic mass unit is a unit of mass while electron volt is a unit of energy. The equivalence between them can be obtained from Einstein's mass-energy relation. According to Einstein, the energy equivalent of a certain mass 'm' is E = mc2 where 97 c is the speed of light in free space. Now, 1 amu of mass is 1.6605402 x 10 kg. Therefore its energy equivalent is given by 2 £ = m (in kg) c2 (ms-1) 2 = 1.6605402 xlO-27 (kg) ^2.99792458 xl08j m2s~2 - 1.6605402 x 8.98755179 x 10~27 x 1016 kg m2s~2 E = 1.6605402 x 8.98755179 x 10"" J . ' 1.6605402 x 8.98755179 xlO"11 ' r , „ t'nnt„no 1n-19 ,1 i.e., E =--— eV v 1 eV = 1.60217738 x10 J 1.60217738 x 10 L J = 9.314 xlO8 eV E = 931.4 x 106 eV i.e., E ~ 931 MeV v 106 eV = 1 Mevj Thus, energy equivalent of 1 amu of mass is nearly 931 MeV. Note : The atomic mass scale is a new mass scale adopted by international union of pure and applied physics in 1960 to measure atomic and nuclear masses. It is defined in terms of the mass of 6C12 atom which is the most abundant and stable isotope of carbon. This scale is also called isotopic mass scale. As a result, mass of an atom or that of the nucleus of an atom is called its isotopic mass. Mass defect Consider the nucleus of an atom of atomic number Z and mass number .4, lzXA], It contains Z protons and (A - Z) neutrons. If mp and mn are the free state rest masses of a proton and a neutron respectively then the rest mass of the nucleus is [Z (m ) + (A - Z) mn). It is found that the rest mass M of a nucleus as obtained from experiments is always less than the total mass of its constituent particles. This is found to be true for all stable nuclei. The difference between the mass of the nucleus and the sum of the free state rest masses of its constituent particles is known as mass defect, denoted by Am. Thus, mass defect of a nucleus is Am = [z (mp) +(A-Z) #n„] - M Binding energy It is found that there is a certain mass defect ("missing" mass) associated with every stable nucleus. The theoretical explanation for this 'missing' mass is based on Einstein's mass-energy relation. If A m be the mass defect of the nucleus of an atom of atomic number Z and mass number A then the energy released during the formation of the nucleus is = (A m) c2. It is called the binding energy of the nucleus. This means that the total energy of all the nucleons in their bound state (when inside the nucleus) is. less than the total energy of the same nucleons in their free state. Hence, binding energy of a nucleus can also be defined as the total energy to be supplied to break the nucleus completely into its constituent particles. Consider the lightest stable atom containing more than one nucleon namely, the deuteron. It has one proton and one neutron in the nucleus. Hence, the mass of H2 is [mp + mn] = 1.007825 + 1.008665 = 2.01649 u. The experimental mass of ,//2 atom is found to be 2.014102 u. This shows that the mass defect of XH2 is Am = 2.016490 - 2.014102 = 0.002238 u. The energy equivalent of this mass is the binding energy of jH2 i.e., Eb =(Am)c2 =(0.002238 u) (931.4) MeV/u = 2.224 MeV This shows that energy needed to break a deuterium nucleus into a neutron and a proton is 2.224 MeV and experiments have proved this result.

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