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Chapter 31

Nuclear Physics and Radioactivity

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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:

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ZA X

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Examples:

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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:

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(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

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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

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−ΔN

Δtwhere N represents the number of nucleii present.

For a random process, the activity is proportional to N:

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ΔN

Δt= −λN

This gives (by integration)

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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

⎛

⎝ ⎜ ⎞

⎠ ⎟

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b) Dating ancient rocks

Age equation: