Lecture 1Lecture 1&& 2 2

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• Calculate the mass defect and the Calculate the mass defect and the binding energy per nucleon for a binding energy per nucleon for a particular isotope.particular isotope.

• Define and apply the concepts of Define and apply the concepts of mass number, atomic number, and mass number, atomic number, and isotopes.isotopes.

All of matter is composed of at least three fundamental particles (approximations):

All of matter is composed of at least three fundamental particles (approximations):

ParticleParticle Fig.Fig. SymSym MassMass ChargeCharge SizeSize

The mass of the proton and neutron are The mass of the proton and neutron are close, but they are about 1840 times the close, but they are about 1840 times the mass of an electron.mass of an electron.

Electron Electron ee-- 9.11 x 10 9.11 x 10-31-31 kg kg -1.6 x 10-1.6 x 10-19 -19 C C Proton Proton pp 1.673 x 101.673 x 10-27-27 kg +1.6 x 10 kg +1.6 x 10-19 -19 C 3 C 3

fmfmNeutron Neutron nn 1.675 x 101.675 x 10--2727 kg kg 0 0 3 fm3 fm

Beryllium AtomBeryllium Atom

Compacted Compacted nucleus:nucleus:4 protons4 protons5 5

neutronsneutrons

Since atom is Since atom is electri-cally electri-cally neutral, there neutral, there must be 4 must be 4 electrons.electrons.4 4 electronselectrons

A A nucleonnucleon is a general term to denote a is a general term to denote a nuclear particle - that is, either a proton or nuclear particle - that is, either a proton or a neutron.a neutron.The The atomic number atomic number ZZ of an element is equal of an element is equal to the number of protons in the nucleus of to the number of protons in the nucleus of that element.that element.The The mass number mass number AA of an element is equal of an element is equal to the total number of nucleons (protons + to the total number of nucleons (protons + neutrons).neutrons).The mass number A of any element is equal to the sum of the atomic number Z and the number of neutrons N :

A = N + Z

A convenient way of describing an element A convenient way of describing an element is by giving its mass number and its atomic is by giving its mass number and its atomic number, along with the chemical symbol for number, along with the chemical symbol for that element.that element.

A convenient way of describing an element A convenient way of describing an element is by giving its mass number and its atomic is by giving its mass number and its atomic number, along with the chemical symbol for number, along with the chemical symbol for that element.that element.

A Mass numberZ Atomic numberX Symbol

A Mass numberZ Atomic numberX Symbol

For example, consider beryllium (Be):

94Be

Lithium AtomLithium Atom

N = A – Z = N = A – Z = 7 - 7 - 33

A = A = 7; Z = 3; 7; Z = 3; NN = ?= ?

Protons: Z = 3Protons: Z = 3

neutrons: neutrons: NN = 4 = 4

Electrons: Electrons: Same Same as Zas Z

73 Li73 Li

Example 1: Describe the nucleus of a lithium atom which has a mass number of 7 and an atomic number of 3.

IsotopesIsotopes are atoms that have the same are atoms that have the same number of protons (number of protons (ZZ11= Z= Z22), but a ), but a different number of neutrons (N). (different number of neutrons (N). (AA11 A A22))

Helium - 4Helium - 4

42He

Helium - 3Helium - 3

32He Isotopes Isotopes

of of heliumhelium

Because of the existence of so Because of the existence of so many isotopes, the term many isotopes, the term elementelement is is sometimes confusing. The term sometimes confusing. The term nuclidenuclide is better. is better.A nuclide is an atom that has a definite mass number A and Z-number. A list of nuclides will include isotopes.The following are best described as The following are best described as

nuclides:nuclides:32He

42He

126C

136C

One One atomic mass unitatomic mass unit (1 u)(1 u) is equal to is equal to one-twelfth of the mass of the most one-twelfth of the mass of the most abundant form of the carbon atom--abundant form of the carbon atom--carbon-12carbon-12..Atomic mass unit: 1 u = 1.6606 x 10-

27 kgCommon atomic masses:

Proton: 1.007276 u

Neutron: 1.008665 u

Electron: 0.00055 u

Hydrogen: 1.007825 u

Electron: Electron: 0.00055 u0.00055 u

= = 11.00930511.009305

The mass of the nucleus is the atomic The mass of the nucleus is the atomic mass less the mass of Z = 5 electrons:mass less the mass of Z = 5 electrons:

Mass = 11.009305 u – 5(0.00055 Mass = 11.009305 u – 5(0.00055 u)u)1 boron nucleus = 11.00656 1 boron nucleus = 11.00656

uu

m = 1.83 x 10-26 kg

m = 1.83 x 10-26 kg

Exampe 2: The average atomic mass of Boron-11 is 11.009305 u. What is the mass of the nucleus of one boron atom in kg?

2 8; 3 x 10 m/sE mc c 2 8; 3 x 10 m/sE mc c Recall Einstein’s equivalency formula for m Recall Einstein’s equivalency formula for m

and E:and E:

The energy of a mass of 1 u can be The energy of a mass of 1 u can be found:found:

EE = (1 u) = (1 u)cc22 = = (1.66 x 10(1.66 x 10-27 -27 kg)(3 x 10kg)(3 x 1088 m/s)m/s)22

E = 1.49 x 10-10 J OrOr E = 931.5 MeV

When When converting amu converting amu

to energy:to energy:

2 MeVu931.5 c 2 MeV

u931.5 c

EE = = mcmc22 = = (1.00726(1.00726 u)(931.5 u)(931.5 MeV/u)MeV/u)

Proton: E = 938.3 MeVProton: E = 938.3 MeV

Similar conversions show Similar conversions show other rest mass energies:other rest mass energies:

Electron: Electron: EE = 0.511 = 0.511 MeVMeVElectron: Electron: EE = 0.511 = 0.511 MeVMeV

Neutron: E = 939.6 MeVNeutron: E = 939.6 MeV

Example 3: What is the rest mass energy of a proton (1.007276 u)?

The Size of NucleiThe Size of NucleiThe size and structure of nuclei were first investigated in The size and structure of nuclei were first investigated in the scattering experiments of Rutherford,the scattering experiments of Rutherford, Using the Using the principle of conservation of energy, Rutherford found an principle of conservation of energy, Rutherford found an expression for how close an alpha particle moving directly expression for how close an alpha particle moving directly toward the nucleus can come to the nucleus before being toward the nucleus can come to the nucleus before being turned around by Coulomb repulsion.turned around by Coulomb repulsion. In such a head-on In such a head-on collision If we equate the initial kinetic energy of the collision If we equate the initial kinetic energy of the alpha particle to the maximum electrical potential energy alpha particle to the maximum electrical potential energy of the system (alpha particle plus target nucleus), we of the system (alpha particle plus target nucleus), we have:have:

Nuclear Size and MassNuclear Size and Mass

The radii of most nuclei is given by the The radii of most nuclei is given by the equation:equation:

R = RR = R00AA1/31/3

R: radius of the nucleus, A: mass number, RR: radius of the nucleus, A: mass number, R00= 1.2x10-= 1.2x10-1515mm

The mass of a nucleus is given by:The mass of a nucleus is given by:

m = A um = A u

u = 1.66…x10u = 1.66…x10-27-27kg (unified mass unit)kg (unified mass unit)

Exercise :Estimate the radius of a uranium-235 nucleus.

Answer: 7.41 x 10-15 m

Nuclear DensityNuclear Density5656

2626Fe is the most abundant isotope of iron (91.7%)Fe is the most abundant isotope of iron (91.7%)

R = 1.2x10-R = 1.2x10-1515m (56)m (56)1/31/3= 4.6x10= 4.6x10-15-15mm

  

ρ= m / V = 2.3x10ρ= m / V = 2.3x101717kg/mkg/m33

The Nuclear ForceThe force that binds together protons and neutrons inside the nucleus is called the Nuclear Force

Some characteristics of the nuclear force are:• Attractive forces at distance between nucleons equal

0.4fm (in spite of coulomb's force).•It is very short range ≈10-15m •It does not depend on charge, Fn-p = Fn-n = Fp-p (Isobars

or Mirror nuclei).•It is much stronger than the electric force•It is saturated (nucleons interact only with near neighbors)•It favors formation of pairs of nucleons with opposite spins.We do not have a simple equation to describe the nuclear force

The The mass defectmass defect is the difference is the difference between the rest mass of a nucleus between the rest mass of a nucleus

and the sum of the rest masses of its and the sum of the rest masses of its constituent nucleons.constituent nucleons.

The The mass defectmass defect is the difference is the difference between the rest mass of a nucleus between the rest mass of a nucleus

and the sum of the rest masses of its and the sum of the rest masses of its constituent nucleons.constituent nucleons.

The whole is less than the sum of the The whole is less than the sum of the parts!parts! Consider the carbon-12 atom Consider the carbon-12 atom

(12.00000 u):(12.00000 u):Nuclear mass = Mass of atom – Electron Nuclear mass = Mass of atom – Electron

masses = 12.00000 u – masses = 12.00000 u – 6(0.00055 u) = 11.996706 6(0.00055 u) = 11.996706 uuThe The nucleusnucleus of the carbon-12 atom has this of the carbon-12 atom has this

mass.mass. (Continued . . .)(Continued . . .)

Mass of carbon-12 nucleus: Mass of carbon-12 nucleus: 11.99670611.996706

Proton: 1.007276 Proton: 1.007276 uu

Neutron: 1.008665 Neutron: 1.008665 uu

The nucleus contains 6 protons and 6 The nucleus contains 6 protons and 6 neutrons:neutrons:

6 p = 6(1.007276 u) = 6.043656 u6 p = 6(1.007276 u) = 6.043656 u

6 n = 6(1.008665 u) = 6.051990 u6 n = 6(1.008665 u) = 6.051990 u

Total mass of parts: = Total mass of parts: = 12.095646 u12.095646 uMass defect Mass defect mmDD = 12.095646 u – = 12.095646 u –

11.996706 u11.996706 umD = 0.098940 u

mD = 0.098940 u

The The binding energy binding energy EEBB of a nucleus is of a nucleus is the energy required to separate a the energy required to separate a nucleus into its constituent parts.nucleus into its constituent parts.

The The binding energy binding energy EEBB of a nucleus is of a nucleus is the energy required to separate a the energy required to separate a nucleus into its constituent parts.nucleus into its constituent parts.

EB = mDc2 where c2 = 931.5 MeV/u

The binding energy for the carbon-12 The binding energy for the carbon-12 example is:example is:

EEBB = (= (0.098940 u)(931.5 MeV/u)

EB = 92.2 MeVBinding EB for C-12:

An important way of comparing the nuclei An important way of comparing the nuclei of atoms is finding their binding energy per of atoms is finding their binding energy per

nucleon:nucleon:

An important way of comparing the nuclei An important way of comparing the nuclei of atoms is finding their binding energy per of atoms is finding their binding energy per

nucleon:nucleon:

Binding energy per

nucleon

MeV =

nucleonBE

A

For our C-12 example A = 12 and:For our C-12 example A = 12 and:

MeVnucleon

92.2 MeV7.68

12BE

A MeV

nucleon

92.2 MeV7.68

12BE

A

The following formula is useful for mass The following formula is useful for mass defect:defect:

D H nm Zm Nm M D H nm Zm Nm M Mass Mass

defectdefect mmDD

mmHH = 1.007825 u; m = 1.007825 u; mnn = 1.008665 = 1.008665 uuZ is atomic number; N is neutron Z is atomic number; N is neutron number; M is mass of atom (including number; M is mass of atom (including

electrons).electrons).By using the mass of the hydrogen atom,

you avoid the necessity of subtracting electron masses.

By using the mass of the hydrogen atom, you avoid the necessity of subtracting

electron masses.

D H nm Zm Nm M D H nm Zm Nm M Mass Mass

defectdefect mmDD

ZmZmHH = (2)(1.007825 u) = 2.015650 u = (2)(1.007825 u) = 2.015650 u

NmNmnn = (2)(1.008665 u) = 2.017330 u = (2)(1.008665 u) = 2.017330 u

MM = 4.002603 u (From nuclide tables) = 4.002603 u (From nuclide tables)

mmDD = (2.015650 u + 2.017330 u) - = (2.015650 u + 2.017330 u) - 4.002603 u 4.002603 u

mD = 0.030377 u

mD = 0.030377 u

EB = mDc2 where c2 = 931.5 MeV/u

EB = mDc2 where c2 = 931.5 MeV/u

EEBB = = (0.030377 u)(931.5 MeV/u) = (0.030377 u)(931.5 MeV/u) = 28.3 28.3 MeVMeV

MeVnucleon

28.3 MeV7.07

4BE

A MeV

nucleon

28.3 MeV7.07

4BE

A

A total of 28.3 MeV is required To tear apart the nucleons from the He-4 atom. A total of 28.3 MeV is required To tear

apart the nucleons from the He-4 atom.

Since there are four nucleons, we find Since there are four nucleons, we find thatthat

Binding energy per nucleon- determines stability- for 4He, BE = 28 MeV so BE/nucleon = 7 MeV

increase in nearest neighbors

increase in Coulomb repulsion dominates

Fusion Fission

Curve shows that Curve shows that EEBB increases with increases with AA and and peaks at peaks at AA = 60 = 60 – about 8 Mev per nucleon. – about 8 Mev per nucleon. Heavier nuclei are less stable. Heavier nuclei are less stable.

Green region is for most stable Green region is for most stable atoms.atoms.

For heavier nuclei, energy is released when For heavier nuclei, energy is released when they break up (they break up (fissionfission). For lighter nuclei, ). For lighter nuclei, energy is released when they fuse together energy is released when they fuse together ((fusionfusion).).

Atomic number Z

Neu

tron

nu

mb

er

N

Stable Stable nucleinuclei

Z = Z = NN

20 40 60 80 100

40

100

140

20

60

80

120

Nuclear particles Nuclear particles are held together are held together by a nuclear strong by a nuclear strong force.force.

Nuclear particles Nuclear particles are held together are held together by a nuclear strong by a nuclear strong force.force.A stable nucleus A stable nucleus remains forever, but remains forever, but as the ratio of as the ratio of N/ZN/Z gets larger, the gets larger, the atoms decay.atoms decay.Elements with Z > Elements with Z > 82 are all unstable.82 are all unstable.Elements with Z > Elements with Z > 82 are all unstable.82 are all unstable.

 

Isotopes are atoms having(a) Same number of protons but different number of neutrons(b) Same number of neutrons but different number of protons(c) Same number of protons and neutrons(d) None of the above