Nuclear Physics & Society (2009) 1 Radioactivity Three main types, all release energy through the relation Q= E= mc 2 –A–Alpha ( ) decay Emission.

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Radioactivity

• Three main types, all release energy through the relation Q=E=mc2

– Alpha () decay• Emission of a 4He nucleus (i.e. 2 protons and 2 neutrons).

– Beta () decay • Changes proton to neutron (or vice versa).

– Gamma () decay• Protons and neutrons re-arrange to lower energy configuration in

same nucleus.

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Proton Drip Line

Neutron Drip Line

Super Heavies

Fewer than 300 nuclei

For a give fixed A (isobar), we have

different combinations of Z and N.

e.g., A=Z+N=137 can be from

Z=56, N=81 →137Ba81 ; or

Z=55, N=82 → 137Cs82 (see later)…

Mass Parabolas and Radioactive Decays

A = constant

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Another View of the Nuclear Landscape

Nuclear masses plotted versus N and Zfor the light nuclei

Remember E = mc2

The valley represents the nuclei with the lowesttotal energy.

The nuclei up on the sides of the valley are unstable and willdecay successively until theyreach the bottom and hencestability.

Our raw materials for nuclearPhysics are the atomic nuclei atthe bottom of the valley-there are283 stable or long-lived isotopes we can find in the Earth’s crust oratmosphere

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Radioactivity: Part 1- Beta decay

• E=mc2 determines if nucleus decays by beta-decay.

• Nuclei try to get to their lowest energy-mass, i.e., lowest mass proton-neutron combination for a given A(=Z+N).

• Beta decay changes a proton to neutron or vice versa.

• The effect is to ‘transmute’ from one chemical element to another, changing Z and N but keeping A fixed.

• An example of beta-decay is 137Cs82 → 137Ba81 + +

• Aside: can leave daughter nucleus in excited state, which decays by Gamma () decay

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A=N+Z = fixed

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Example of a mass parabola

Mass energy

(mc2)

A=N+Z=125

p →

n + + +

125Xe : Z=54; N=71

125Cs : Z=55; N=70

125I : Z=53; N=72

125Ba : Z=56; N=69

125Te : Z=52; N=73

STABLE ISOBAR

FOR A=125

125In : Z=49

125Sn : Z=50;

125Sb : Z=51;

n →

p + - +

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Radioactivity

– Alpha () decay (cont.)• Emission of a 4He nucleus (i.e. 2 protons and 2 neutrons).

• energies are characteristic to the specific decay.

• Can leave daughter in an excited state which can then decay by..

– Gamma () decay• Protons and neutrons re-arrange to lower nuclear energy state.

• No change in N and Z, ray emitted (high energy/ frequency ‘light’).

• Typical energies ~100 – 2000 keV

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Nuclear Alpha Decay

0+

4+

6+

2+

0+

0.0046 %

0.035 %

25.0 %

74.0 %

24296Cm

23894Pu Q g.s. = 6.216 MeV

t 1/2= 163 d

= ( -1 )

Alpha decay can leave daughterIn excited states which can thendecay by (characteristic) gammaemission.

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Alpha-decay of 210Po: Another example of

E=mc2

4He (Z=2) (N=2)

206Pb (Z=82) (N=124)

210Po (Z=84) (N=126)

Q = energy release = [ M(210Po) – { M(206Pb) + M(4He)} ] c2 ~ 5.5 MeV ~1 pJ.

i.e., around 0.003% of total 210Po mass released kinetic energy in decay.

(rare occasion, 206Pb can be left in an excited state….803 keV above the ground state.)

Total energy released shared as KINETIC energy between 206Pb and 4He.

Cons. of linear momentum means most of the energy (5.3 MeV) goes to 4He.

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Energy Release in Decay: A Sum!!• Use E = mc2 to calculate particle energies in 210-Polonium

decay……

• Recall, 1u = 1 atomic mass unit = 931.5 MeV/c2 (=1.66x10-27 kg)

• M(210Po) = 209.982848u• M(206Pb) = 205.974440u• M(4He) = 4.002603u

• E=Q = { M(210Po) –[ M(206Pb) + M(4He)] } c2

• E=Q = [209.982848 – (205.97444 + 4.002603)] x 931.5 MeV

E=Q = 0.005805 x 931.5 MeV = 5.407 MeV

BUT energy is split between emitted particle and 206Pb such that

T = Q / [1 + M/ M206Pb ] = 5.407 / [1+4/206] =5.407 / 1.019 = 5.3 MeV

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spectroscopyis very useful indetermining whichalpha emitters arepresent…..

Particularly useful in identifying minoractinides from nuclearwaste (such as 241Am)

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Radiation in our Environment

We are all constantly subject to irradiation mainly from natural sources.

There are three main sources of such radiation.

a) Primordial -around since the creation of the Earth ( 4.5 x 109years) 235,8U ( and daughters including 210Po), 232Th or 40K (+ 87Rb, 138La and others....)

b) Cosmogenic – from interaction of Cosmic rays with Earth and atmosphere. 14C, 7Be formed from cosmic ray interactions. Cosmic rays are mostly protons.

c) Produced or enhanced by human activity. Medical or dental X-rays; 137Cs (product from nuclear fission, 239Pu, 241Am, 239Pu from weapons fallout

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Radioactive species in the body

Isotope Average amount by weight Activity

U-Uranium 90μg 1.1Bq

Th-Thorium 30 μg 0.11Bq

40K 17mg 4.4 kBq

Ra 31pg 1.1Bq

14C 22ng 3.7kBq

3H-tritium 0.06pg 23Bq

Po-Polonium 0.2pg 37Bq

Some variation- for example smokers have 4-5 times more Po.

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Q210Pb) = 5.41 MeVE = 5.30 MeV E(206Pb) = 0.11 MeVT1/2 = 138 days.

‘218Po =Radium A’

‘218At =Radium B’

C

D

E

210Po=Radium ‘F’ Radon

=‘Emanation’

‘Radium’

C’

C’’

The Natural Decay Chain for 238U

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