Post on 20-Jan-2016
transcript
© J Parkinson1
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Mass Defect
The difference between the mass
of the atom and the sum of the masses
of its parts is called the mass defect (m).
Careful measurements show that the mass of a particular atom is always slightly less than the sum of the masses of the individual protons, neutrons and electrons of which the atom consists.
e.g. a helium nucleus consists of 2 protons and 2 neutrons
2 protons & 2 neutrons Helium atom
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mp = mass of a proton (1.007277 amu)
mn = mass of a neutron (1.008665 amu)
me = mass of an electron (0.000548597 amu)
1 atomic mass unit ( amu ) = 1.661 x 10-27 kg
m = [ Z(mp + me) + (A-Z)mn ] - matom
1 amu = 1/12 of the mass of an atom of Carbon-12
Z = PROTON NUMBER (ATOMIC NUMBER) AND ALSO EQUALS THE NUMBER OF ELECTRONS
A = NUCLEON NUMBER (MASS NUMBER) AND ALSO EQUALS THE NUMBER OF
PROTONS + NEUTRONS
THE NUMBER OF NEUTRONS MUST BE A - Z
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For Helium the mass defect will be
m = [ 2(1.007277 + 0.000548597 ) + 2 x 1.008665 ] – 4.021435 = 0. 0115462 amu
0. 0115462 amu x 1.661 x 10-27 = 1. 91782 x 10-29 kg
Using Einstein’s E = mc2, this is equivalent to a loss of energy given by
E = 1.91782 x 10-29 x (3 x 108)2 Joules = 1.726 x 10-12 J
This figure is the BINDING ENERGY of the Helium nucleus.
THE BINDING ENERGY of a nucleus is defined as the energy which must be input to separate all of its protons and neutrons.
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Binding Energies are usually expressed in MeV 1 amu = 931.3 MeV
The binding energy of the Helium nucleus is therefore
0. 0115462 amu x 931.3 = 10.75 MeV
To compare the stabilities of different nuclei,
Binding Energies PER NUCLEON in the nucleus are compared.
For Helium this will be 10.75 ÷ 4 = 2.69 MeV per nucleon
The higher the binding energy per nucleon, the greater the stability of the nucleus
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NUCLEON NUMBER
BINDING ENERGY
Per NUCLEON
2H
238U
56Fe8.8 MeV
Iron is the most stable nucleus
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BE/A
A
ALTERNATIVELYBE/A
A0 250
9.0
As energy is lost when nuclei are synthesised
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FISSION
IRON
Heavy nuclei may increase their stability by Nuclear Fission
Light nuclei may increase their stability by Nuclear Fusion
FUSION
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As atomic mass increases, the neutron to proton ratio for stable nuclei increases because proton-proton repulsion becomes significant, as the number of protons increases. Cohesive nuclear forces arise form neutrons, so the neutron to proton ratio must increase for heavier elements.
Belt of StabilityBelt of Stability
Proton number, Z
Ne
utr
on
nu
mb
er,
N =
A -
Z
N = Z
Bel
t of S
tabl
e Is
otop
es
For helium He- 4 the N:P ratio is 1 : 1
For uranium U- 238 the ratio is 1 : 1.6
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Nuclear Fission is the fragmentation of heavy nuclei to form lighter, more stable ones.
The Fission of U - 235
U23592
energyofMeVnKrBanU 1803 10
9436
13956
10
23592
This is only one of several fissures that are possible.
On average 2.5 neutrons are released
© J Parkinson11
Neutrons released in the fission of 235U can induce three more fissions, then nine, and so on leading to a chain reaction.
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Critical mass is the mass required for the chain reaction to become self-sustaining.
neutron
Some neutrons may :
Cause more fission
Get lost
Be absorbed by an atom
lost
For a chain reaction to be self sustaining, every fission must produce at least one more
neutron that will initiate further fissions.
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Nuclear Reactor
Pressure vessel and biological shield
Graphite Moderator slows neutrons to thermal energies [1 MeV]
Fuel Rods remain in reactor until spent
Coolant [CO2 or H2O] to extract energy
To heat exchanger where steam is produced
Boron control rods absorb neutrons to control the reaction