Chapter 31
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Transcript of Chapter 31
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1. Nuclear Structure
a) Proton- positive charge- mass 1.673 x 10-27 kg ≈ 1 u
b) Neutron- discovered by Chadwick (student of Rutherford)- hypothesized to account for mass of atom- discovered with scattering experiments- zero charge- mass 1.675 x 10-27 kg ≈ 1 u - mass of neutron ≈ mass of proton + mass of electron- neutron can eject electron to form proton, but it’s not a
proton and an electron
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c) Nucleon - constituent of nucleus (neutron or proton)
d) NomenclatureA - number of nucleons (atomic mass number)Z - number of protonsN - number of neutrons
A = Z + N
Symbol for nucleus of chemical element X:
€
ZA X
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Examples:
€
hydrogen nucleus : 11H
helium 4 : 24He
neutron : 01n
electron : -10e
aluminum : 1327 Al ≡ 27Al
Since Z determines the element (X), only AX is needed.
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e) Atomic mass unit, uDefine: Mass of 12C = 12 uThen,
1 u = 1.66 x 10-27 kg = 931.5 MeV/c2
mp = 1.00727 u
mn = 1.008665 u
In chemistry and biology, 1 dalton (Da) = 1 u
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f) Isotopes; nuclei with the same Z, different N
e.g. 35Cl, 37Cl (65%, 35%), 12C, 13C, 14C (99%, 1%, 0.01%)
g) Nuclear size and density
Close-packed- constant density- Volume proportional to atomic number (A)- Since V = 4/3 πr3, A prop. to r3
- r prop. A1/3
- r ≈ (1.2 x 10-15 m) A1/3 = 1.2 fm A1/3
- density of neutron star = 100 million tonne/cm3
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2. Nuclear force and stability
a) Strong nuclear force- one of the fundamental forces - holds protons together in spite of Coulomb
repulsion- short range: ~ fm (zero for longer range)
- only adjacent nucleons interact
- acts equally between n-p, n-n, p-p
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b) Symmetry
c) Coulomb repulsion
- Pauli exclusion principle: N=Z gives maximum stability considering only nuclear force
-long range; all protons interact (only adjacent nucleons feel nuclear force)
- repulsion increases with size -- neutron excess needed for stability
- above Z = 83 (Bi) stability not possible; larger elements decay emitting radioactivity
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3. Mass defect and binding energy
a) Binding energyenergy required to separate constituents of nucleus
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€
B.E . + mc 2 = m1c2 + m2c
2 + ...= mic2∑
b) Mass defect
From special relativity, adding energy increases mass:
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B.E . = mi − m∑{ }c 2
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B.E . = Δmc 2
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Δm = mass defect = constituents - composite
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Example: 4He (alpha particle)
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Δm = (2mp + 2mn − m 4 He) = 0.0304u
Compare ionization potential for H atom: 13.6 eV
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BEα = Δmc 2 = (0.0304u)c2
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=0.0304931.5MeV
c 2
⎛ ⎝
⎞ ⎠c 2
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=28.3 MeV
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c) Atomic electrons
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4 He + B.E . → 2n +21H
-Masses usually tabulated for neutral atoms (including atomic electrons)- Can use atomic masses if electrons balance:
€
(4 He + 2e) + B.E .→ 2n + 2(1H + e)
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d) Binding energy per nucleon
increase in nearest neighbors
increase in Coulomb repulsion dominates
- determines stability- for 4He, BE = 28 MeV so BE/nucleon = 7 MeV
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Energy released
Fusion Fission
For a given number of nucleons,- if BE/nucleon increases- mass defect increases- total mass decreases- energy released
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Potential energy diagram for nucleons: fusion releases energy
Fusion:
Energy (high temperature in the sun) required to push nuclei together against the Coulomb force.
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Potential energy diagram for two halves of a large nucleus: fission releases energy
Fission:
May occur spontaneously, or be induced by neutron bombardment
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4. Radioactivity
- spontaneous decay of nucleus- releases energy to achieve higher BE/nucleon- mass of parent > mass of products
18a) - decay
- ejection of 4He nucleus
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ZA P →Z −2
A −4 D+ 24He
- transmutation: element changes
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E = Δmc 2
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=(mp − md − mα )c 2
- Energy released (KE of , daughter, energy of photon)
Use atomic masses for P, D, 4He (electrons balance):
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E = (mp(atom) − md (atom) − m 4 He(atom))c 2
For 238U, 234Th, 4He, E = 4.3 MeV
-decay
19b) - decay- ejection of electron
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ZA P →Z +1
A D+−10β
- transmutation
- Energy released, as KE of electron
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E = Δmc 2
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=(mp − md − mβ )c 2
Use atomic masses for P, D, and add one electron mass:
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E = (mp(atom) − md (atom))c2
- decay
For 234Th and 234Pa, E = 0.27 MeV
- governed by weak nuclear force
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- ejection of positron
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ZA P →Z −1
A D++10β
- electron capture
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ZA P+−1
0β →Z −1A D
Other modes of beta decay
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c) - decay
- emission of a photon
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ZA P∗ →Z
A P + γ- no transmutation
- accompanies - decay, fission, neutron decay
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d) Decay series- sequential decays to an eventual stable nucleus
QuickTime™ and aTIFF (Uncompressed) decompressor
are needed to see this picture.
- 4 separate series (A can only change by 4)
238U -> 206Pb235U -> 207Pb
232Th -> 208Pb
237Np -> 209Bi (not obs’d)
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e) Neutrino, - postulated by Pauli in 1930 to account for missing energy in -decay
€
ZA P →Z −1
A D++10β + ν
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ZA P →Z +1
A D+−10β + ν
- observed in 1956
- mass ~ zero (< ~ eV) (standard model predicts non-zero mass)
- could account for missing mass in universe
- zero charge
- interacts only by weak nuclear force (difficult to detect)
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5. Radioactive decay rate; activity
a) ActivityActivity is the number of decays per unit time, or
€
−ΔN
Δtwhere N represents the number of nucleii present.
For a random process, the activity is proportional to N:
€
ΔN
Δt= −λN
This gives (by integration)
€
N = N0e−λ t
where N0 is the number of nuclei at t = 0.
Units: 1 Bq (becquerel) = 1 decay/s1 Ci (curie) = 3.7 x 1010Bq (activity of 1 g radium)
is the decay constant
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b) Half-lifeExponential decay:
For a given time interval, the fractional decrease in N is always the same:
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N1
N2
=e−λ t1
e−λ t2= e−λ ( t1 −t2 )
Define half-life as the time for activityto reduce by 1/2:
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1
2= e−λ (T1/2 )
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T1/ 2 =ln2
λ=
.693
λ
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Using
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e−λ t = 2−λ t / ln 2
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N = N02−λt / ln 2
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N = N0
1
2 ⎛ ⎝
⎞ ⎠
t /T1/2
the exponential can be expressed
so
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N = N0
1
2 ⎛ ⎝
⎞ ⎠
number of half -lives
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6. Radioactive dating
a) Carbon dating- based on the reaction:
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14 C →14 N +β T1/2 = 5730 years
- 14C/12C ratio constant in atmosphere due to cosmic rays
- living organisms ingest atmospheric carbon; dead matter doesn’t
- ratio of 14C/12C in matter gives time since death
Equilibrium ratio: 1/8.3 x 1011
==> 1 g C contains 6 x 1010 atoms of 14C
==> Activity of 1g C (at eq’m) = 0.23 Bq = A0
==> Activity of 1g C (time t after death) = A= A0e-t
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t =−1
λln
A
A0
⎛
⎝ ⎜ ⎞
⎠ ⎟