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					                                           Nuclear Physics                                                IB 12

  Nuclide: a particular type of nucleus

  Nucleon: a proton or a neutron

  Atomic number (Z) (proton number): number of protons in nucleus

  Mass number (A) (nucleon number): number of protons + neutrons

  Neutron number (N): number of neutrons in nucleus (N = A – Z)

  Isotopes: nuclei with same number of protons but different numbers of neutrons

  Unified atomic mass unit (u): 1/12th the mass of a carbon-12 nucleus                             1 u = 1.661 x 10-27 kg
                                                                                                   1 u = 1 g/mol
  Atomic mass ≈ A * u                                                                              1 u = 931.5 MeV/c2

        Z     X                   56
                                  26   Fe                  27
                                                           13     Al                12
                                                                                    6    C               14
                                                                                                         6    C
  Atomic Number                      26                          13                      6                     6

    Mass Number                      56                          27                      12                   14

  Neutron Number                     30                          14                      6                     8

    Atomic Mass                     56 u                         27 u                  12 u                   14 u

     Molar Mass                     56 g                         27 g                  12 g                   14 g

  How big are atomic nuclei? 10-15m – 10-14 m                           How do we know that neutrons exist?
                                                                        By the existence of isotopes
  How do we know this? Alpha-particle
  scattering experiments (Geiger-Marsden)                               How do we know that isotopes exist?
                                                                        By measurements in the mass spectrometer
  ET = ET
  EK = Ee
  ½ mv2 = qV
  ½ mv2 = q(kQ/r)

                                                                        v = E/B
Note that most nuclei have approximately the same . . .density
= 2 x 1017 kg/m3 = 1014 times denser than water                         r = mv/qB
                                          Nuclear Stability                                   IB 12
What interactions exist in the nucleus?

       1. Gravitational: (long range) attractive but very weak/negligible

       2. Coulomb or Electromagnetic: (long range) repulsive and very strong between protons

       3. Strong nuclear force: (short range) attractive and strongest – between any two nucleons

       4. Weak nuclear force: (short range) involved in radioactive decay

 Why are some nuclei stable while others are not?

   The Coulomb force is a long-range force which means that every
   proton in the nucleus repels every other proton. The strong
   nuclear force is an attractive force between any two nucleons
   (protons and/or neutrons). This force is very strong but is short
   range (10-15 m) which means it only acts between a nucleon and
   its nearest neighbors. At this range, it is stronger than the
   Coulomb repulsion and is what holds the nucleus together.

   Neutrons in the nucleus play a dual role in keeping it stable. They
   provide for the strong force of attraction, through the exchange of
   gluons with their nearest neighbors, and they act to separate
   protons to reduce the Coulomb repulsion.

   Each dot in the plot at right represents a stable nuclide and the
   shape is known as the “band (or valley) of stability.” With few
   exceptions, the naturally occurring stable nuclei have a number N of
   neutrons that equals or exceeds the number Z of protons. For small
   nuclei (Z < 20), number of neutrons tends to equal number of
   protons (N = Z).

   As more protons are added, the Coulomb repulsion rises faster than the strong force of attraction
   since the Coulomb force acts throughout the entire nucleus but the strong force only acts among
   nearby nucleons. Therefore, more neutrons are needed for each extra proton to keep the nucleus
   together. Thus, for large nuclei (Z > 20), there are more neutrons than protons (N > Z).
   After Z = 83 (Bismuth), adding extra neutrons is no longer able to counteract the Coulomb repulsion
   and the nuclei become unstable and decay in various ways.

   Nuclei above (to the left of) the band of stability have too many neutrons and tend to decay by alpha
   or beta-minus (electron) emission, both of which reduce the number of neutrons in the nucleus.

   Nuclei below (to the right of) the band of stability have too few neutrons and tend to decay by beta-
   plus (positron) emission which increases the number of neutrons in the nucleus.

                                                 Binding Energy                                               IB 12

 The total mass of a nucleus is always less than the sum of the masses its nucleons. Because mass is another
manifestation of energy, another way of saying this is the total energy of the nucleus is less than the combined
                                      energy of the separated nucleons.

Mass defect (mass deficit) (Δm)

Difference between the mass of the nucleus and the sum of the masses of
its individual nucleons

Nuclear binding energy (ΔE)
1. energy released when a nuclide is assembled from its individual                         mnucleus + Δm = mnucleons
                                                                                                E = (  m )c
2. energy required when nucleus is separated into its individual

Different nuclei have different total binding energies. As a general trend, as the atomic number
increases . . . the total binding energy for the nucleus increases.

                     Electric Charge        Electric Charge           Rest Mass            Rest Mass             Rest Mass
                            (e)                   (C)                   (kg)                  (u)                (MeV/c2)
   Proton                    +1               +1.60 x 10-19          1.673 x 10-27          1.007276                    938
   Neutron                    0                    0                 1.675 x 10-27          1.008665                    940
   Electron                  -1               -1.60 x 10-19          9.110 x 10-31          0.000549                0.511

1. The most abundant isotope of helium has a 24He nucleus whose mass is 6.6447 × 10-27 kg. For this nucleus, find the
   mass defect and the binding energy.

           helium: 0.050404 x 10-27 kg
                 4.53636 x 10-12 J

2. Calculate the binding energy and mass defect for 816O whose measured mass is 15.994915 u.

                                                                                     Oxygen: 0.132613 u
                                                                                        123.5 MeV                   3
                                         Binding Energy per Nucleon                                                IB 12

     To see how the nuclear binding energy varies from nucleus to nucleus, it is useful to compare the binding
                  energy for each nucleus on a per-nucleon basis, as shown in the graph below.

                                                                                                       Your Turn

a)    This graph is used to compare the energy states of different nuclides and to determine what nuclear reactions are energetically
      feasible. As binding energy per nucleon increases so does the stability of the nucleus. Higher binding energies represent
      lower energy states since more energy was released when the nucleus was assembled.

b) Binding energy per nucleon increases up to a peak at 2656Fe then decreases, so 2656Fe is the most stable nuclide. Most nuclides
   have a binding energy per nucleon of about 8 MeV. Lighter nuclei are held less tightly than heavier nuclei.

c)    Nuclear reactions, both natural (radioactive decay) and artificial/induced (fission, fusion, bombardments) occur if they
      increase the binding energy per nucleon ratio. Fusion occurs for light nuclei (below 2656Fe) and fission occurs for heavy nuclei
      (above 2656Fe).

d) For both natural and induced nuclear reactions, the total rest mass of the products is less than the total rest mass of the
   reactants since energy is released in the reaction. Also, the products are in a lower energy state since energy was released in
   the reaction and so the products have a greater binding energy per nucleon than the reactants.

1. Use the graph above to estimate the total binding energy of an oxygen-16 nucleus. 8 MeV x 16 = 128 MeV

                                              Types of Nuclear Reactions

1) Artificial (Induced) Transmutation: A nucleus is bombarded with a nucleon, an alpha particle or another small nucleus,
resulting in a nuclide with a different proton number (a different element).

2) Nuclear Fusion: Two light nuclei combine to form a more massive nucleus with the release of energy.

3) Nuclear Fission: A heavy nucleus splits into two smaller nuclei of roughly equal mass with the release of energy.

4) Natural Radioactivity: When an unstable (radioactive) nucleus disintegrates spontaneously, the nucleus emits a particle
of small mass and/or a photon.

Release of energy in nuclear reactions:              m = m + Δm
Energy is usually released in the form of kinetic energy for the products.
Binding energy per nucleon: greater for product nuclei than for original nuclei since energy is released.
                                      Radioactive Decay Reactions                                                IB 12
                                                   Alpha Decay
Alpha particle: helium nucleus, α, 24He

Example reaction:         226
                          88     Ra 86 Rn 4 He  energy

General equation:         A
                          Z   X Z 4 Y 2 He  energy

Where does the kinetic energy come from? Rest mass of nucleons

Result: nucleus is in a more stable state with higher binding energy and higher BE per nucleon
since it released energy

1. A radium nucleus, initially at rest, decays by the emission of an alpha particle into radon in the reaction described above.
    The mass of 88226Ra is 226.025402 u and the mass of 86222Rn is 222.017571 u and the mass of the alpha particle is
    4.002602 u.
                                                                                         Δm = 0.005229 u
    a) Calculate the energy released in this decay.

                                                                                              4.87 MeV

    b) Compare the momenta, speeds, and kinetic energies of the two particles produced by this reaction.

    c) If the kinetic energy of the alpha particle is                       d) Calculate the recoil speed of the radon nucleus.
        4.77 MeV, calculate its speed.                                          Use momentum – no need to change units

                                                         Beta Decay                                    IB 12
      Beta-minus particle: electron, β , -1 e     0

      Beta-plus particle: positron, β , +10e

      Consider the following two “mysterious” results of beta decay:

          a)   Observe the before and after picture of beta decay.
               What’s wrong?

          b) Inspect the graph of kinetic energy carried away by
             the beta particles. Why are so few beta particles
             leaving with the majority of the kinetic energy?
             Where did this missing kinetic energy go?

      Conclusion: there is third particle involved with beta decay that carries away some KE
      and momentum – virtually undetectable

      Neutrino and anti-neutrino: fundamental particles - no charge – very small mass

                    Beta-minus decay                                                 Beta-plus decay

Example reaction:                                                      Example reaction:

6     C 14 N 01 e 0   energy
         7          0
                                                                       7    N 12 C 01 e 0   energy
                                                                               6          0

General equation:                                                      General equation:

Z   X Z 1 Y 01 e 0   energy
                    0
                                                                       Z   X Z 1 Y 01 e 0   energy
                                                                                           0

How does this happen? Weak force                                       How does this happen? Weak force

0    n 1 p 01 e 0   energy
        1         0
                                                                       1   p 1 n 01 e 0   energy
                                                                              0    

                                                        Gamma Decay
    Gamma particle: high energy photon, 

    Example reaction:        12
                             6    C* 12 C    energy
                                                                                                       Before Decay

    General equation:         A
                              Z   X * Z X    energy

                                                                                                       After Decay
    Where does the photon (energy) come from? Rest mass of nucleons

                   Energy Spectra of Radiation                                                                            IB 12
   The nucleus itself, like the atom as a whole, is a quantum system           Importance: discrete energy spectra give evidence
   with allowed states and discrete energy levels. The nucleus can be          for nuclear energy levels
   in any one of a number of discrete allowed excited states or in its
   lowest energy relaxed state. When it transitions between a higher
   energy level and a lower one, it emits energy in the form of alpha,
   beta, or gamma radiation. When an alpha particle or a gamma
   photon is emitted from the nucleus, only discrete energies are                                                                   Gamma
                                                                                Alpha spectra            Beta spectra
   observed. These discrete energy spectra give evidence that a                                                                     spectra
   nucleus has energy levels. (However, the spectrum of energies
   emitted as beta articles is continuous due to its sharing the energy             Discrete              continuous                discrete
   with a neutrino or antineutrino in any proportion.)

                                                       Ionizing Radiation
Ionizing Radiation – As this radiation passes through materials, it “knocks off” electrons from neutral atoms thereby
creating an ion pair: free electrons and a positive ion. This ionizing property allows the radiation to be detected but
is also dangerous since it can lead to mutations in biologically important molecules in cells, such as DNA.

                                α                    β                      γ
                                                Electron or
       Particle           helium nucleus                           high-energy photon
  Penetration ability           low               medium                   high
                      Sheet of paper; a
  Material needed to                              1 mm of
                     few centimeters of                               10 cm of lead
      absorb it                                  aluminum
   Path length in air        a few cm        less than 1 meter      effectively infinite

                                      Detection of Radiation: the Geiger-Muller tube (Geiger counter)

                             The Geiger counter consists of a gas-filled metal cylinder. The α, β, or γ rays enter the cylinder through a thin
                             window at one end. Gamma rays can also penetrate directly through the metal. A wire electrode runs along the
                             center of the tube and is kept at a high positive voltage (1000-3000 V) relative to the outer cylinder.

                             When a high-energy particle or photon enters the cylinder, it collides with and ionizes a gas molecule. The
                             electron produced from the gas molecule accelerates toward the positive wire, ionizing other molecules in its
                             path. Additional electrons are formed, and an avalanche of electrons rushes toward the wire, leading to a pulse of
                             current through the resistor R. This pulse can be counted or made to produce a "click" in a loudspeaker. The
                             number of counts or clicks is related to the number of disintegrations that produced the particles or photons.

                                                         Biological Effects of Ionizing Radiation

   Alpha and beta particles have energies typically measured in MeV. To ionize an atom requires about 10 eV so each particle can
   potentially ionize 105 atoms before they run out of energy. When radiation ionizes atoms that are part of a living cell, it can affect
   the ability of the cell to carry out its function or even cause the cell wall to rupture. In minor cases, the effect is similar to a burn. If
   a large number of cells that are part of a vital organ are affected then this can lead to death. Alternatively, instead of causing the cell
   to die, the damage done by ionizing radiation might just prevent cells from dividing and reproducing. Or, it could be the cause of the
   transformation of the cell into a malignant form. If these malignant cells continue to grow then this is called cancer.

   The amount of harm that radiation can cause is dependent on the number and energy of the particles. When a gamma photon is
   absorbed, the whole photon is absorbed so one photon can ionize only one atom. However, the emitted electron has so much energy
   that it can ionize further atoms, leading to damage similar to that caused by alpha and beta particles.

   On a positive note, rapidly diving cancer cells are very susceptible to the effects of radiation and are more easily killed than normal
   cells. The controlled use of the radiations associated with radioactivity is of great benefit in the treatment of cancerous tumors.
                                Mathematical Description of Radioactive Decay                                     IB 12
Radioactive decay:

   1) Random process: It cannot be predicted when a particular nucleus will decay, only the
      probability that it will decay.

   2) Spontaneous process: It is not affected by external conditions. For example, changing the
      pressure or temperature of a sample will not affect the decay process.

   3) Rate of decay decreases exponentially with time: Any amount of radioactive nuclei will reduce
      to half its initial amount in a constant time, independent of the initial amount.

 Half-life (T1/2)                                                                                                Units:
 - the time taken for ½ of the radioactive nuclides in a sample to decay                                         = s or hr or
                                                                                                                   d or yr

 - the time taken for the activity of a sample to decrease to ½ of its initial value

 N0 = number of radioactive nuclei originally present

 N = number of radioactive nuclei present at any one time                                             Your Turn

                                                                                      Radioactive tritium has a half-life of about 12
                                                                                           years. Complete the graph below.
                         Radioactive X → stable Y + particle

                         Show how amount of daughter Y mirrors X

 A nuclide X has a half-life of 10 s. On decay the stable nuclide Y is formed. Initially a sample contains only atoms of X. After
 what time will 87.5% of the atoms in the sample have decayed into nuclide Y? 30 s

                                                         Activity                                                 IB 12

Activity (A) – the number of radioactive disintegrations per unit time (decay rate)

                                                       Units:             decays/time = s-1 or hr-1or d-1 or yr-1
                                   t                 Standard
                                                                            Becquerel (Bq)

                                                                            1 Bq = 1 decay per second

1. A sample originally contains 8.0 x 10 12 radioactive nuclei and has a half-life of 5.0 seconds. Calculate the activity of
    the sample and its half-life after:

      a) 5.0 seconds                               b) 10. seconds                            c) 15 seconds

         8.0 x 1011 Bq                                 6.0 x 1011 Bq                             4.7 x 1011 Bq
         5.0 s                                         5.0 s                                     5.0 s

2. Samples of two nuclides X and Y initially contain the same number of radioactive nuclei, but the half-life of nuclide
    X is greater than the half-life of nuclide Y. Compare the initial activities of the two samples.

                                                                           Activity of sample Y is greater


         A N
                                            The Radioactive Decay Law: The rate at which radioactive nuclei in a
                                            sample decay (the activity) is proportional to the number of radioactive
                                            nuclei present in the sample at any one time.

         A  N                             [As the number of radioactive nuclei decreases, so does the average
                                            rate of decay (the activity).]
    Initial Activity

         A0   N0                          The initial activity (A0) is directly related to the number of radioactive
                                            nuclei originally present (N0) in the sample.

  3. The isotope Francium-224 has a half-life of 20 minutes. A sample of the isotope has an initial activity of 800
    disintegrations per second. What is the approximate activity of the sample after 1 hour?

                                                                                  N = 1/8 N0
                                                                                So A = 1/8 A0
                                                                           A = 100 disintegrations s-1

                                                                                                                  IB 12
Decay constant (λ)                                                                                              Units:
– constant of proportionality between the decay rate (activity) and the                                         inverse time
  number of radioactive nuclei present.
                                                                                                                = s-1 or hr-1or d-1
                                                                                                                or yr-1

- probability of decay of a particular nuclei per unit time.

Deriving the Radioactive Decay Law                                Relating the Decay Constant and Half-life

                                                                 At t  T1/ 2 then N  N 0
                                                                                          1                                     1
                   N                                                                     2           1      T    
               A     N
                                                                                                      2   e 1/ 2 
                   t                                          The decay equation becomes
                                                                                                                    
                dN                                                        N  N0e t
                     N                                                                                2  eT1/ 2
                                                                         N  N eT1/ 2
                                                                           0   0
                                                                                                       ln  2  e 1/ 2 
                                                                                                                      
       A solution to this equation
                                                                                                          ln 2T1/ 2
        is an exponential function
                                                                                                            ln 2 0.693
                of the form                                                                                    
                                                                                                            T1/ 2 T1/ 2
                  N  N 0 e  t                                                                                ln 2 0.693
                                                                                                      T1/ 2        
                                                                                                                           
                       t
           A = Aoe             N 0 e  t

1. The half-life of a certain radioactive isotope is 2.0 minutes. A particular nucleus of this isotope has not decayed within
    a time interval of 2.0 minutes. What is the probability of it decaying in:

    a) the next two minutes                     b) the next one minute                 c) the next second

2. A sample of a radioactive isotope X has the same initial activity as a sample of the isotope Y. The sample of X
    contains twice the number of atoms as the sample of Y. If the half-life of X is TX then the half-life of Y is 0.5 TX

                                                                                                                    IB 12
3. The half-life of a radioactive isotope is 10 days. Calculate the fraction of the sample that will be left after 15 days. 35%

4. The half-life of a radioactive substance is 10 days. Initially, there are 2.00 x 10 26 radioactive nuclei present.
    a) What is the probability of any one particular nucleus decaying?

    b) What is the initial activity?

    c) How many radioactive nuclei are left after 25 days?

    d) What is the activity of the sample after 25 days?

    e) How long will it take for the activity to fall to 1.0 x 1024 dy-1?

                                        Graphs of Radioactive Decay                                             IB 12

      Radioactive nuclei vs. time                          N  N 0 e  t                          Straightening by natural log

                                                     ln N  ln  N 0 e  t 
                                                                                           ln(N0)

                                                   ln N  ln N 0     t
                                                   ln N     t  ln N 0
                                                           y  mx  b

                                                    T1/2 = ln 2/(-slope)                                   Slope = -λ

      Activity of sample vs. time                                                                     Straightening by natural log

                                            Methods of Determining Half-life

If the half-life is short, then readings can be taken of activity versus time using a Geiger counter, for example. Then,

     1.   A graph of activity versus time would give the exponential shape and several values for the half-life could be
          read from the graph and averaged.

     2.   A graph of ln (activity) versus time would be linear and the decay constant can be calculated from the slope.

If the half-life is long, then the activity will be effectively constant over a period of time. If a way could be found to
calculate the number of nuclei present chemically, perhaps using the mass of the sample and Avogadro’s number, then the
activity relation or the decay equation could be used to calculate half-life.

                                                                                                               IB 12
1. Cesium-138 decays into an isotope of barium. Measurements of
  the activity of a particular sample of cesium-138 were taken and
  graphed at right.

  a) Suggest how the data for this graph could have been obtained.
  Geiger counter

  b) Use the graph to estimate the half-life of cesium-138.
  35 minutes

  c) Use the graph to estimate the half-life of the barium isotope.
  Wait until 250/300 minutes when very little cesium left
  90 minutes

2. A 2.0 mg sample of carbon-14 is measured to have an activity of 6.5 x 10 10 Bq.

  a) Use this information to determine the half-life of carbon-14 in years. 5700 years

  b) A student suggests that the half-life can be determined by taking repeated measurements of the activity and
      analyzing the data graphically. Use your answer to part (a) to comment on this method of determining the half-life.

3. The radioactive isotope potassium-40 undergoes beta decay to form the isotope calcium-40 with a half-life
    of 1.3 x 109 yr. A sample of rock contains 10 mg of potassium-40 and 42 mg of calcium-40.

   a) Determine the age of the rock sample.             3.1 x 109 yr

      b) What are some assumptions made in this determination of age?
      No calcium-40 originally present = all potassium-40
      No loss of either isotope in intervening years

                                                Nuclear Fission                                                IB 12

  Nuclear Fission: A heavy nucleus splits into two smaller nuclei of roughly equal mass with the release of energy.

  Nuclear Fusion: Two light nuclei combine to form a more massive nucleus with the release of energy.

  Release of energy in nuclear reactions:
  m = m + Δm
  Energy is usually released in the form of . . .
  kinetic energy for the products.

  Binding energy per nucleon: greater for
  product nuclei than for original nuclei since
  energy is released.

                   One Common Fission Reaction

        92    U 1 n 92 U *  X  Y  neutrons

There are about 90 different daughter nuclei (X and Y) that can be formed.
Here is a typical example:

           92    U 1 n 141 Ba 92 Kr  31 n
                    0    56      36       0
1. Estimate the amount of energy released when a uranium nucleus fissions.
         Use plot above
         Original: 7.5 MeV per nucleon
         Daughters: 8.5 MeV
         Energy difference: 1.0 MeV per nucleon x 235 nucleons
         = 235 MeV as KE of daughters and neutrons

2. A neutron collides with a nucleus of plutonium and the following fission reaction occurs. Determine the number of neutrons
    produced and calculate the amount of energy released.

   94        Pu 1 n 140 Ba 96 Sr 
                 0    56      38
                                                                                                m = m + Δm
   94     Pu = 239.052157 u                                                                 0.19383 u = 180 MeV
     38 Sr = 95.921750 u
   56     Ba = 139.910581 u
        0 n = 1.008665 u                                                                                          14
                                        Chain Reactions                                        IB 12

Chain Reaction – neutrons released from one fission reaction go on to initiate further reactions

Uncontrolled nuclear fission: nuclear weapons

Controlled nuclear fission: nuclear power production

   1) some material absorbs excess neutrons before striking nucleus

                                                                                     Uncontrolled Chain Reaction
   2) leaving only one neutron from each reaction to produce
      another reaction

Critical Mass: if mass of uranium is too small, too many
neutrons escape without causing further fission in uranium so
the reaction cannot be sustained

                                                                                       Controlled Chain Reaction

Thermal Neutron: low-energy neutron (≈1eV) that favors fission reactions – energy comparable to
gas particles at normal temperatures

Naturally Occurring Isotopes of Uranium:

   1) Uranium-238: most abundant, 99.3%, very small probability of fissioning when it captures a
      neutron, not used for fuel, more likely to capture high energy neutron than low energy one

   2) Uranium-235: 0.3%, 500 times greater probability of fissioning when captures a neutron but must
      be a low-energy (thermal) neutron, used for fuel

Fuel Enrichment: process of increasing proportion of uranium-235 in a sample of uranium

       1) formation of gaseous uranium (uranium hexafluoride) from uranium ores

       2) Separated in gas centrifuges by spinning – heavier U-238 moves to outside

       3) increases proportion of U-235 to about 3% to be used as fuel in nuclear reactors

         Advantage: more uranium is available for fission and reaction can be sustained

         Disadvantage: enriched fuel can be used in the manufacture of nuclear weapons –
         threat to world peace – 85% = weapons grade
                                       Nuclear Reactors                                       IB 12

Most nuclear reactors: thermal fission reactor using uranium-235 as fuel

   Fuel Rods: enriched solid uranium

   Moderator: material (water, graphite) used to slow down high-energy neutrons emitted from fission
   reactions to thermal levels for use in further fission reactions to sustain the chain reaction - slow
   neutrons by collisions

   Control Rods: inserted between fuel rods – made of neutron-absorbing cadmium or boron -
   used to control reactor temperature to prevent overheating – lowered if too many
   neutrons/reactions and excess thermal neutrons are absorbed

   Heat Exchanger: hot fluid circulating around fuel rods (primary loop) is fed into tank of water –
   heat is transferred to water and makes steam – steam expands adiabatically against fan blades of
   turbines and turns a magnet is a coil of wire to generate electricity

Neutron Capture and Plutonium-239                                                                IB 12
    Uranium-238 is a non-fissionable isotope but is considered “fertile”

         92    U 1 n 92 U 93 Np 01 e 
                       239   239
                                                                           Neutron capture and
                                                                           Beta-minus decay

          93    Np 94 Pu 01 e 
                                                                           Beta-minus decay

          94    Pu 1 n 140 Ba 96 Sr  31 n
                    0    56      38       0                                reaction

       Advantage: plutonium-239 used as fuel in “breeder reactors”

       Disadvantage: plutonium-239 used in nuclear weapons

                Safety Issues and Risks in the Production of Nuclear Power

 Uranium Mining:
  open-cast mining: environmental damage, radioactive waste rock (tailings)
   underground mining: release of radon gas (need ventilation), radioactive rock dangerous for workers,
 radioactive waste rock (tailings)
 leaching: solvents pumped underground to dissolve uranium and then pumped back out –
 contamination of groundwater

 Thermal Meltdown: overheating and melting of fuel rods – may be caused by malfunction in cooling
 system or pressure vessel – overheating may cause pressure vessel to burst sending radioactive material
 and steam into atmosphere (as in Chernobyl, Ukraine 1986) – hot material may melt through floor –
 “China syndrome” as in Three Mile Island – limited by containment vessel and containment building

 Nuclear Waste:
 Low-level waste: radioactive material from mining, enrichment and operation of plant – must be
 disposed of – left untouched or encased in concrete

 High-level waste: disposal of spent fuel rods- some isotopes have ½ lives of thousands of years –
 plutonium 240,000 years
     1) stored under water at reactor site for several years to cool of then sealed in steel cylinders,
        buried underground
     2) reprocessed to remove any plutonium and useful uranium, remaining isotopes have shorter
        ½ lives and long-term storage need is reduced

 Nuclear Weapons Manufacture:
 Enrichment technology could be used to make weapons grade uranium (85%) rather than fuel
 grade (3%)
 Plutonium is most used isotope in nuclear weapons and can be gotten from reprocessing spent
 fuel rods
                                                                                                               IB 12
1. Suppose the average power consumption for a household is 500 W per day. Estimate the amount of uranium-235 that would have
    to undergo fission to supply the household with electrical energy for a year. Assume that for each fission, 200 MeV is released.

2. A fission reaction taking place in a nuclear power station might be    235
                                                                          92    U 1 n 141 Ba 92 Kr  31 n
                                                                                   0    56      36       0
    Estimate the initial amount of uranium-235 needed to operate a 600
    MW reactor for one year assuming 40% efficiency and 200 MeV
    released for each fission reaction.

                                                 Nuclear Fusion                                                IB 12

Nuclear Fusion: Two light nuclei combine to form a more massive nucleus with the release of energy.

1. Write the reaction equation for the fusion reaction shown at right.

            1   H 1 H 2 He 1 n
                   3    4

2. Calculate how much energy is released in this fusion reaction.

                                 m = m + Δm                                                            2
                                                                                                      1   H (deuterium, 2.0141 u)
                                                                                                          1 H (tritium, 3.0161 u)
                                0.0189 u = 17.6 MeV                                                            4
                                                                                                              2 He (4.0026 u)
                                                                                                             neutron (1.0087 u)

3. Calculate the energy released per nucleon and compare this with a fission reaction.

                     5 nucleons react = 17.6 MeV / 5 = 3.5 MeV
                     Compared with ≈1.0 MeV for fission

Important occurrence of fusion: main source of Sun’s energy – fusion of hydrogen to helium

Suggested Mechanism: proton-proton cycle

                 1   H 1 H 1 H 1 e 
                             2    0

                 1   H 1 H 3 He  

           Then either:                                                   Or:

                1   H 3 He 4 He 1 e  
                       2     2
                                   0                                     3
                                                                         2   He 3 He 4 He 1 H 1 H
                                                                                 2     2     1    1

                                                     Fusion Reactor                                          IB 12

  Plasma: fuel for reactor – high energy ionized gas (electrons and nuclei are
  separate) – if energy is high enough (hot enough), nuclei can collide fast enough to
  overcome Coulomb repulsion and fuse together

  Magnetic confinement: charged particles are contained via magnetic fields –
  travel in a circle in a doughnut shaped ring (tokamak)

  Heating Plasma: accelerate nuclei by means of magnetic fields and forces = high
  temperatures (high kinetic energies)

  Problems with current fusion technology: maintaining and confining very high-density and
  high-temperature plasmas – very difficult to do – uses more energy input than output – not
  commercially efficient

                                  Artificial (Induced) Transmutation

Artificial (Induced) Transmutation: A nucleus is bombarded with a nucleon, an alpha particle or another small nucleus,
resulting in a nuclide with a different proton number (a different element).

Requirement: the bombarding particle must have sufficient kinetic energy to overcome the Coulomb

1. In 1919, Ernest Rutherford discovered that when nitrogen gas is bombarded with alpha particles, oxygen and protons are
    produced. Complete the equation for this reaction.

                             7    N 2 He 17 O 1 H
                                           8     1

2. Neutron bombardment of lithium can produce the radioactive isotope of hydrogen known as tritium. Complete the reaction.

                             3   Li 1 n 1 H 4 He

  NOTE: isotopes produced

  Importance: artificial isotopes produced are used in medical tests and therapies