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Set 57 Problem number 21


Problem

The table below shows masses, in atomic units, of various neutral isotopes of several elements.  If an atomic mass unit is 1.660502 * 10^-27 kg, if a proton has mass 1.007276 amu and a neutron mass 1.008665 amu, and if a neutral hydrogen atom has mass 1.007825 amu, what is the mass defect of the isotope in the third row of column # 10?  What is the mass defect per nucleon for this isotope?

What is the binding energy of this atom, in electron volts?  What is the binding energy per nucleon?

Properties of Selected Nuclei

particle or atom

cobalt

nickel

iron

gold

mercury

lead

bismuth

polonium

radium

thorium

atomic number

27

28

26

79

80

82

83

84

88

90

mass (amu)

59.933822

59.93079

56.9354

197.9682

197.9668

205.9744

208.980383

208.9824

226.0254

230.0331

mass (amu)

56.936292

209.984105

Solution

The element in Column 10 has atomic number 88, which is equal to the number of charged particles in the nucleus--i.e., to the number of protons.  We therefore have 88 in the nucleus.

The atomic mass is 226.0254, representing the number of atomic mass units in a neutral atom.  This mass is very close to a whole number of atomic mass units.   If we round this number off we obtain the number of nucleons--i.e., the total number of protons and neutrons--in the nucleus. 

So we see that there are 226 nucleons in the nucleus.  Since 88 of these are protons, it is easy to see that we have 226 - 88 = 138 neutrons in the nucleus.

The atom consists of its neutrons, protons and electrons, one electron for each proton.  A basic hydrogen atom consists of a single proton and a single electron.   Therefore if we add the masses of all the neutrons, plus the mass of a hydrogen atom for each proton, we will get the total unbound mass of all the particles that make up the atom:

This is somewhat greater than the mass of the actual atom, which is 226.0254 amu.   The difference is

and represents the energy that binds the particles of the atom together.

Since there are 226 nucleons in the nucleus, the mass defect per nucleon is

The binding energy is the energy equivalent of the mass defect:

The binding energy per nucleon is therefore

This binding energy is easiest to understand in units of electron volts:

=  ( 76.76 * 10^-13 J / nucleon) * (1 eV / {1.6 * 10^-19 eV/Joule} )

= 47.98 * 10^6 eV

=  47.98 * 10^6 MeV.

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