nuclear physics and radioactivity. vocabulary alpha particle - positively charged particle...
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Vocabulary
alpha particle - positively charged particle consisting of two protons and two neutrons. (Helium nucleus)
atomic mass number (A) - the number of protons and neutrons in the nucleus of an atom.
atomic mass unit - the unit of mass equal to 1/12 the mass of a carbon-12 nucleus; the atomic mass rounded to the nearest whole number is called the mass number.
atomic number (Z) - the number of protons in the nucleus of an atom.beta particle - high speed electron emitted from a radioactive element
when a neutron. decays into a protonbinding energy – the energy required to completely separate the nucleus
into its individual protons and neutrons.element - a substance made of only one kind of atom.isotope - a form of an element which has a particular number of neutrons,
that is, has the same atomic number but a different mass number than the other elements which occupy the same place on the periodic table.
Vocabulary
mass defect - the mass equivalent of the binding energy in the nucleus of an atom by E = mc2
neutron - an electrically neutral subatomic particle found in the nucleus of an atom
nuclear reaction - any process in the nucleus of an atom that causes the number of protons and/or neutrons to change
nucleons - protons or a neutronsstrong nuclear force - the force that binds protons and neutrons together
in the nucleus of an atomtransmutation - the changing of one element into another by a loss of gain
of one or more protons
Equations and symbols
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ΔE = Δm( )c 2
1u =1.66x10−27 kg = 931MeVc−2 XAZ
where ΔE = binding energy of the nucleusΔm = mass defect of the nucleusc = speed of light = 3 x 108 m/su = atomic mass unitX = element symbolA = atomic mass number (number of protons and neutrons)Z = atomic number (number of protons)
H11
n10
e01
Particle Symbol Relative mass Charge Location
proton1 +1 nucleus
neutron1 0 nucleus
electron or e- 0 -1electron orbitals
around the nucleus
Find the number of protons, electrons and neutrons in a neutral atom of iron.
Fe5626
# protons
# protons + #neutrons
26 protons
30 neutrons
neutral → #protons = # electrons = 26
M2M1
Before After
M 5
M4
M3
Eo Ef=<Mo Mf=>
Eo = Ef + (Mo-Mf)c2
Eo = Ef + (Δm)c2
MASS is transferred to ENERGY
Calculate the binding energy of Nitrogen. The atomic mass of Nitrogen is 14.003074 u.
Nucleon Mass (u)
Proton 1.00782
Neutron 1.00866
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714N
7 protons7 neutrons
Mass of individual protons and neutrons7(1.00782u) + 7(1.00866u) = 14.11536u
Δm = 14.11536u – 14.003074u = .112286u
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.112286u ×931MeVc−2
1u=105MeVc−2
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E = Δm( )c 2 = 105MeVc−2( )c
2 =105MeV
Transmutation of nitrogen into carbon
714N0
1n 614C
Nuclide/particle Mass (u)
14.003074
neutron 1.008664
14.003241
proton 1.007825
14 N
14C
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11H
Mass of products14.003074u + 1.008664u = 15.011738u
Mass of reactants14.003241u + 1.007825u = 15.011066u
Δm = 15.011738u – 15.011066u = .000672u
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.000672u ×931MeVc−2
1u= 0.626MeVc−2
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E = Δm( )c 2 = 0.626MeVc−2( )c
2 = 0.626MeV
Nuclear fusion – The joining of two small nuclei to form one large nucleus. The mass of the smaller nuclei is greater than the mass of the large nucleus. High temperatures are required for fusion. Example: the sun
Nuclear fission – Splitting a large nucleus into two smaller nuclei. The mass of the large nucleus is greater than the two smaller nuclei. Example: Nuclear power plant
In both cases the mass of the products is less than the mass of the reactants, which results in a mass defect. The mass is converted to energy according to the equation E = (Δm)c2
Binding energy per nucleon vs. mass number
Binding energyPer nucleon
(MeV)
Mass number
Iron (Fe) staple
fusi
onFission
Fusion Examples:
Nuclide Mass (u)
1.007825
2.014101
3.016049
3.016029
4.002603
1.008664
1 H
3He
3 H
2 H
4Hen1
0
12H1
3H 24He0
1n
12H1
2H 23He0
1n
12H1
2H 13H1
1p
12H2
3He 24He1
1p
Nuclide Mass (u)
1.007825
2.014101
3.016049
3.016029
4.002603
1.008664
1 H
3He
3 H
2 H
4Hen1
0
12H1
3H 24He0
1n
Mass of reactants2.014101u + 3.016049u = 5.03015u
Mass of products4.002603u + 1.008664u = 5.011267u
Δm = 5.03015u – 5.011267u = .018883u
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.018883u ×931MeVc−2
1u=17.5MeVc−2
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E = Δm( )c 2 = 17.5MeVc−2( )c
2 =17.5MeV
Fission Example: absorbs a neutron and splits into and , Write the equation for the nuclear reaction and calculate the energy released in this reaction.
U23692 Mo100
42
Sn12650
Z symbol A Mass (u)
92 U 236 236.045563
42 Mo 100 99.907476
50 Sn 126 125.907653
Fission Example: absorbs a neutron and splits into and , Write the equation for the nuclear reaction and calculate the energy released in this reaction.
U23692 Mo100
42
Sn12650
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92236U +0
1 n→42100 Mo +50
126 Sn + neutrons
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92236U +0
1 n→42100 Mo +50
126 Sn + ? 01n( )
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92236U +0
1 n→42100 Mo +50
126 Sn +11 01n( )
Z symbol A Mass (u)
92 U 236 236.045563
42 Mo 100 99.907476
50 Sn 126 125.907653
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92236U +0
1 n→42100 Mo +50
126 Sn +11 01n( )
Mass of products236.045563u + 1.008664u = 237.054227
Mass of reactants125.907653u + 99.907476u + 11(1.008664u) = 236.910433u
Δm = 237.054227u – 236.910433u = .143794u
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.143794u ×931MeVc−2
1u=134MeVc−2
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E = Δm( )c 2 = 134 MeVc−2( )c
2 =134MeV
Radioactivity – Particles are randomly emitted from an unstable nucleus in order to become more staple. These are the different particles emitted.
β+positron
High Energy
Photonsgamma
β-Beta
αAlpha
symbolName
He42
e01
e01
Nuclear Equations:
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90234Th→ 91
234 Pa + ?
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23490Th→ 91
234 Pa+−10e
-1
0
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23490Th→ 91
234 Pa + Β−
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+ν
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ν = anti - neutrino
Bet
a de
cay
Nuclear Equations:
nPAl 10
3015
2713? 4
2
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24He+13
27Al→1530 P+0
1n
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+1327Al→15
30 P+01n
Transmutation