Nuclear Energy

Mass Defect

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Introduction • Nuclear power stations tap into the energy stored

inside an atom’s nucleus.• Huge amounts of energy can be released.• But where does it come from and how can we get

at it?

Learning Outcomes:• At the end of this lesson you will be able to:1. Explain Mass Defect2. Calculate the energy stored by an atomic nucleus

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Missing mass – Mass Defect• A helium nucleus is made up of 2 protons and 2

neutrons.• When the mass of two protons + mass of 2

neutrons is compared with mass of helium nucleus – something very odd is seen.

• The mass of the nucleus is less than the total mass of the individual particles that it contains

• True for all nuclei containing more than one nucleon.

• Missing mass is known as mass difference or mass defect.

• Masses involved are very small – measured in atomic mass units u, where:

• 1u = 1.6605 x 10-27 kg

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Mass defectMass

(atomic mass unit)

Proton 1.00728 u

Neutron 1.00867 u

Helium nucleus 4.00151 u

Table 1: Atomic mass units.

Example 1. Using the data in table 1, calculate the mass defect for a helium nucleus in atomic mass units and in kilograms.Mass of 2p + 2n = (2 x 1.00728 u) + (2 x 1.00867 u) = 4.03190 u

Mass defect = 4.03190 u – 4.00151 u = 0.03039 u

Mass defect in kg = 0.03039 x 1.6605 x 10-27 kg = 5.046 x 10-29 kg

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Mass defect• A mass of 5.046 x 10-27 kg may not sound a lot –

but on atomic scale it is significant.• The mass defect is about 3% of the mass of a

proton (or 55 electrons)• It is important to measure nuclear masses

precisely – masses are quoted to 6 significant figures.

Mass and energy equivalence• Splitting the nucleus into individual nucleons

results in increase of total mass.• Where does this extra mass come from?• Splitting the nucleus is very difficult – nucleons

held together by a very strong but short-range nuclear forces.

• Overcoming these forces requires energy5

Mass defect• What happens to this energy?• Energy disappears into the system and mass is

created!• Goes against the conventional conservation laws.

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Albert Einstein

It was Albert Einstein who suggested that mass and energy are equivalent.

He linked mass and energy in his famous equation: E = m . c 2E is the energy equivalent in joules of a mass m in kilograms, c is the velocity of light (3.00 x 108 m s-1)

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Example 2:Calculate the energy equivalent of the mass defect calculated in Example 1.

E = m c2 = 5.046 x 10-29 kg x (3.00 x 108 m s-1)2

= 4.54 x 10-12 J (3 s.f.)

This is the energy needed to separate the helium nucleus into individual protons and neutrons.

42He + 4.54 x 10-12 J → 2p + 2n energy

In any system, the total amount of mass and energy is conserved.

Energy equivalent of 1 u

• Einstein’s equation uses mass in kg and energy in J.• In nuclear physics we are more likely to be working

in atomic mass units (u) and electron-volts (eV)• Need to work out energy equivalent (in eV) of 1 u:• Using Einstein’s equation with a precise value for

the velocity of light:• E = m c2 = 1.6605 x 10-27 kg x (2.9979 x 108 m s-

1)2

= 1.4924 x 10-10 J• But 1 eV = 1.6022 x 10-19 J• Hence E = 1.4924 x 10-10 J / 1.6022 x 10-19

= 9.315 x 108 eV = 931.5 x 106 eV• 1 u = 931.5 MeV

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Example 3:A carbon nucleus has a mass of 11.9967 uHow much energy, in MeV, would ne needed to split it into its 6 protons and 6 neutrons?(mp = 1.00728 u, mn = 1.00867 u)

Mass of 6p + 6 n = (6 x 1.00728 u) + (6 x 1.00867 u)

= 12.0957 uMass defect = 12.0957 u – 11.9967 u = 0.0990 u

Energy equivalent = 0.0990 x 931.5 MeV = 92.2 MeV needed to separate the 12 nucleons.

Question:Calculate the mass defect in u and kg for the following nuclei:(a)Lithium (7

3Li), nuclear mass = 7.014353 u(b)Silicon (28

14Si), nuclear mass = 27.96924 u

Binding energy• Energy needed to separate a nucleus into individual

nucleons is its binding energy.• Also the energy equivalent of the mass defect –

found from E = m c2

• Binding energy indicates the stability of the nucleus.

• Total binding energy is linked to the size of the nucleus.

• The more nucleons there are, the greater the energy needed to separate them all out.

• More useful comparison is the binding energy per nucleon.

• Average energy needed to remove each nucleon from the nucleus.

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Example 4:Use the figures calculated in example 3 to find the binding energy per nucleon for a carbon-12 nucleus.

Carbon-12 contains 12 nucleons (6p + 6n)Total binding energy for carbon-12 = 92.2 MeV

Binding energy per nucleon = 92.2 MeV / 12 = 7.68 MeV

Question For each of the two nuclei in question on slide 9, calculate:(a)Total binding energy in eV(b)The binding energy per nucleon in eV

Binding energy

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Fig 1: Graph of binding energy per nucleon (eV) against nucleon number (A)

Binding energy

• Notice that binding energy per nucleon is typically around 8 MeV

• The most stable nucleus has the highest binding energy per nucleon.

• This is 8.79 MeV for 5626Fe

Radioactive Decay and Binding Energy• Unstable nucleus emits radiation and becomes

more stable.• Daughter nucleus always has a higher binding

energy per nucleon than the parent.• Energy is given out when a nucleus decays• Where does it come from?

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Radioactive Decay and Binding Energy

• Total mass of products is less than mass of parent nucleus.

• Mass difference is released as energy – look at the following examples:Example 5: α-decay

• Thorium-228 decays by α-emission:228

90Th →22488Ra + 4

2α

• Mass of thorium-228 nucleus = 227.97929 u• Mass of radium-224 nucleus + α-particle

= 223.97189 u + 4.00151 u = 227.97240 uMass difference = 227.97929 u – 227.97340 u= 0.00589 u = 5.49 MeV (as 1 u = 931.5 MeV)

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Radioactive Decay and Binding Energy• The surplus energy appears mostly as K.E. of the α-

particle.• Radium nucleus also recoils slightly (momentum is

conserved).Example 6: β-decay

• Aluminium-29 decays by β-emission

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2913Al → 29

14Si + 0-1β + 0

0ν-

Mass of Si-29 nucleus + β-particle + antineutrino= 28.96880 u + 0.000549 u + 0 = 28.969349

u

Mass of aluminium-29 nucleus = 28.97330 u

Mass difference = 28.97330 u – 28.960349 u = 0.003951 u

= 3.68 MeV (as 1 u = 931.5 MeV)

Radioactive Decay and Binding Energy

• Some of this energy is carried away by a γ-ray.• Rest becomes KE of the decay products.

Transmutation and energy• Transmutation is conversion of one element to

another • Radioactive decay is a spontaneous

transmutation with a release of energy.• Some elements can be made by firing very fast

moving charged particles at a stable nucleus – artificial transmutation. (energy must be supplied)

• First achieved by Rutherford, Marsden and Chadwick in 1919.

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Transmutation and energy

• Converted nitrogen into oxygen by bombarding it with α-particles.14

7N + 42α → 17

8O + 11H

• Here mass of products is greater.• Energy must be supplied to make this reaction

happen and balance the nuclear equation• Energy comes from K.E. of the α-particle. Modern

use of artificial transmutation is the production of medical radioisotopes.

Questions1. The decay of radium into radon is

22688Ra → 222

86Rn + 42

α

calculate the energy released in eV17

Questions

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2. A possible reaction for the fission of uranium is:235

92U + 10n → 146

57La + 8735Br + 31

0n Calculate the energy released in the reaction in

MeV.3. Calculate the minimum energy needed in MeV to

make the following reaction happen:60

28Ni + 42α → 63

30Zn + 10nNuclear masses: radium-226 = 225.9771 u

radon-222 = 221.9703 u uranium-235 = 234.9934 u lanthanum-146 = 145.8684 u bromine-87 = 86.9153 u nickel-60 = 59.9153 u zinc-63 = 62.9205 u