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NUCLEAR REACTIONS - FISSION AND
FUSION

Problem 1
When U undergoes fission, about 0.1 % of the
      235
       92


original mass is released as energy.
a) How much energy is released when 1 kg of
  U undergoes fission?
235
 92




b) How much U must undergo fission per day
                235
                 92


   in a nuclear reactor that provides energy to a
   100 MW electric power plant?
c) When coal is burned, about 32.6 MJ/kg of
   heat is released. How many kilograms of
   coal would be consumed per day by a
   conventional coal-powered 100 MW electric
   power plant?

Solution
a) E = mc2 = (0.001 kg)(3 x 108 m/s)2 = 9 * 1013
    J

b) Energy = power x time, so
E = Pt = (108 W) (3600 s/h)(24 h/d) = 8.64 x
    1012 J/day

                        1
Hence the mass of           235
                             92   U   required is
                     12
             8.64 *10 J / day
                     13
                              = 9.6 *10 −2 kg / day = 96 g / day
               9 * 10 J / kg



c) The energy liberated per kg of coal burned is
3.26 x 107 J. Hence the mass of coal required is
     12
8.64 *10 J / day
      7
                 = 2.65 *10 kg / day ,
                     5
                                       which is 265 metric
 3.26 *10 J / kg

tonnes.

The core of the Sun is considered to extend
from the center to about 0.2 solar radii. It has a
density of up to 150,000 kg/m³ (150 times the
density of water on Earth) and a temperature
of close to 13,600,000 kelvins (by contrast, the
surface of the Sun is around 5,800 kelvins).

Problem 2
In the sun and other stars the principal process
that liberates energy is the process of conversion
of hydrogen to helium in a series of nuclear
fusion reactions where positrons (positively
charged electrons) are emitted.
  (a) write the equation for the overall
     process in which 4 protons form a helium
     nucleus (alpha particle) .

                                        2
(b) how much energy is liberated in such a
    process?

Given the masses of H , He and the electrons are
                        1
                        1
                                4
                                2


1.007825, 4.002603, and 0.000549 u.



Solution

(a) the positrons must be given off so that
  charge will be conserved. Hence the overall
  process is
                                   +   +
            1
            1H + H + H + H → He + e + e
                1
                1
                    1
                    1
                        1
                        1
                                    4
                                    2


(b) since helium atoms has only 2 electrons
around its nucleus (2 p+ = 2 e-,ie. x p+ = x e-,
right?), 2 electrons as well as 2 positrons are lost
when each helium atom is formed. The mass
change is therefore
 ∆m = 4mH + 0mn – (mHe + 4me) = (4) (1.007825
       u) – [4.002603 u + (4) (0.000549 u)]
                    = 0.026501 u
    Hence, the amount of energy liberated is
     (0.026501 u) (931 MeV/u) = 24.7 MeV.


                            3
RADIOACTIVITY

Unstable nuclei undergo radioactive decay to
become more stable nuclei. There are 5 types of
radioactive decay (decay events):

                 Decay event      What causes
                                this decay type
 gamma ray      Gamma ray       Nuclear has
                emission        excessive
                reduces energy energy
                of nucleus
 alpha decay    Alpha particle Nucleus is too
                emission        large
                reduces size of
                nucleus
 beta decay     Emission of     Nucleus has
                electron by     too many
                nuclear         neutrons
                neutron         relative to
                changes it to a number of
                proton          protons
 electron       Capture of      Nucleus has
 capture        electron by     too many
                nuclear proton protons

                       4
changes it to a relative to
               neutron         number of
                               neutrons
 positron      Emission of     Nucleus has
 emission      positron by     too many
               nuclear proton protons
               changes it to   relative to
               proton          number of
                               neutrons

Problem 3
What happens to the atomic number and mass
number of a nucleus that
  (a) emits an electron,
  (b) undergoes electron capture,
  (c) emits an alpha particle.

Solution
  (a) Z increases by 1, A is unchanged.
  (b) Z decreases by 1, A is unchanged.
  (c) Z decreases by 2, A decreases by 4.




                      5
Problem 4
How many successive α-decays occur in the
decay of the thorium isotope Th into the lead
                                 228
                                  90


isotope Pb ?
        212
         82




Solution
Each α-decay means a reduction of 2 in atomic
number and 4 in mass number. Here Z decreases
by 8 and A by 16, which means that 4 α-
particles are emitted.

HALF LIFE, t 1/2 = ln 2 / λ

The time required for half of the atoms in any
given quantity of a radioactive isotope to decay
is the half-life of that isotope. Each particular
isotope has its own half-life. For example, the
half-life of 238U is 4.5 billion years. That is, in
4.5 billion years, half of the 238U on Earth will
have decayed into other elements. In another 4.5
billion years, half of the remaining 238U will
have decayed. One fourth of the original
material will remain on Earth after 9 billion
years. The half-life of 14C is 5730 years, thus it
is useful for dating archaeological material.

                         6
Nuclear half-lives range from tiny fractions of a
second to many, many times the age of the
universe. (http://www.lbl.gov/abc/Basic.html).


Problem 5
Tritium is the hydrogen isotope H whose nucleus
                                              3
                                              1


contains two neutrons and one proton. Tritium is
beta-radioactive and emits an electron.
  (a) What does tritium become after
    undergoing beta decay?
  (b) The half-life of tritium is 12.5 year.
    How much of the 1-g sample will remain
    undecayed after 25 years?

Solution
  (a) in the beta decay of nucleus one of the
    neutrons become a proton. Since the atomic
    number A= 2 corresponds to helium, the
    beta decay of tritium is
                     3
                     1   H → 23 He + e −

     and the new atom is         3
                                 2   He   .

  (b) 25 years = 2 x t1/2 , and so ½ x ½ x 1 g = ¼
  g of tritium remains undecayed.

                             7
Problem 6
The half-life of the sodium isotope Na against
                                            24
                                            11


beta decay is 15 h. how long does it take for ⅞th
of a sample of this isotope to decay?

After    ⅞th has decayed, ⅛th is left,and
1 1 1 1
  = × × which is 3 half-lives. Hence the answer
8 2 2 2
is (3)(15) h = 45 h.


Problem 7
The carbon isotope carbon-14 ( C ) called
                                       14
                                        6


radiocarbon is beta-radioactive with a half-life
of 5600 years. Radiocarbon is produced in the
earth’s atmosphere by the action of cosmic rays
on nitrogen atoms, and the carbon dioxide of the
atmosphere contains a small proportion of
radiocarbon as a result. All plants and animals
therefore contain a certain amount of
radiocarbon along the stable isotope of C . When 12
                                                  6


a living thing dies, it stops taking in radiocarbon
and the radiocarbon it already contains decays
steadily. By measuring the ratio between C and        12
                                                       6


 C of the remains of an animal or plant and
14
 6




                         8
comparing it with the ratio of these isotopes in
living organisms, the time that has passed since
the demise of the animal or the plant can be
found.
   (a) How old is a piece of wood from an
     ancient dwelling if its relative radiocarbon
     content is ¼th that of a modern specimen?
   (b) If it is 1/16th that of a modern specimen?

Solution
(a) since ¼ = ½ x ½ , the specimen is two half-
  lives old, which is 11,200 yr old.
(b) since 116 = ½ x ½ x ½ x ½ , the specimen is 4
  half-lives old, which is 22,400 yrs old.




                        9
Problem 8
The rate at which a sample of radioactive
substance decays is called its activity. A unit of
activity is the curie(Ci) where 1 curie =3.7 *
1010 decays/s. if a luminous watch dial contains
5 µCi of the radium isotope Ra , how many
                                       226
                                        88


decays per second occur in it? (This isotope is –
alpha-radioactive).

Solution
Since 1 µCi is 10-6 Ci, the activity of the watch
dial is
( 3.7 ×10 decays / s ) (5 x 10-6 Ci) = 1.85 x 105 decays/s.
      10

             Ci




                            10

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Lesson Nuclear Reactions Radioactivity

  • 1. NUCLEAR REACTIONS - FISSION AND FUSION Problem 1 When U undergoes fission, about 0.1 % of the 235 92 original mass is released as energy. a) How much energy is released when 1 kg of U undergoes fission? 235 92 b) How much U must undergo fission per day 235 92 in a nuclear reactor that provides energy to a 100 MW electric power plant? c) When coal is burned, about 32.6 MJ/kg of heat is released. How many kilograms of coal would be consumed per day by a conventional coal-powered 100 MW electric power plant? Solution a) E = mc2 = (0.001 kg)(3 x 108 m/s)2 = 9 * 1013 J b) Energy = power x time, so E = Pt = (108 W) (3600 s/h)(24 h/d) = 8.64 x 1012 J/day 1
  • 2. Hence the mass of 235 92 U required is 12 8.64 *10 J / day 13 = 9.6 *10 −2 kg / day = 96 g / day 9 * 10 J / kg c) The energy liberated per kg of coal burned is 3.26 x 107 J. Hence the mass of coal required is 12 8.64 *10 J / day 7 = 2.65 *10 kg / day , 5 which is 265 metric 3.26 *10 J / kg tonnes. The core of the Sun is considered to extend from the center to about 0.2 solar radii. It has a density of up to 150,000 kg/m³ (150 times the density of water on Earth) and a temperature of close to 13,600,000 kelvins (by contrast, the surface of the Sun is around 5,800 kelvins). Problem 2 In the sun and other stars the principal process that liberates energy is the process of conversion of hydrogen to helium in a series of nuclear fusion reactions where positrons (positively charged electrons) are emitted. (a) write the equation for the overall process in which 4 protons form a helium nucleus (alpha particle) . 2
  • 3. (b) how much energy is liberated in such a process? Given the masses of H , He and the electrons are 1 1 4 2 1.007825, 4.002603, and 0.000549 u. Solution (a) the positrons must be given off so that charge will be conserved. Hence the overall process is + + 1 1H + H + H + H → He + e + e 1 1 1 1 1 1 4 2 (b) since helium atoms has only 2 electrons around its nucleus (2 p+ = 2 e-,ie. x p+ = x e-, right?), 2 electrons as well as 2 positrons are lost when each helium atom is formed. The mass change is therefore ∆m = 4mH + 0mn – (mHe + 4me) = (4) (1.007825 u) – [4.002603 u + (4) (0.000549 u)] = 0.026501 u Hence, the amount of energy liberated is (0.026501 u) (931 MeV/u) = 24.7 MeV. 3
  • 4. RADIOACTIVITY Unstable nuclei undergo radioactive decay to become more stable nuclei. There are 5 types of radioactive decay (decay events): Decay event What causes this decay type gamma ray Gamma ray Nuclear has emission excessive reduces energy energy of nucleus alpha decay Alpha particle Nucleus is too emission large reduces size of nucleus beta decay Emission of Nucleus has electron by too many nuclear neutrons neutron relative to changes it to a number of proton protons electron Capture of Nucleus has capture electron by too many nuclear proton protons 4
  • 5. changes it to a relative to neutron number of neutrons positron Emission of Nucleus has emission positron by too many nuclear proton protons changes it to relative to proton number of neutrons Problem 3 What happens to the atomic number and mass number of a nucleus that (a) emits an electron, (b) undergoes electron capture, (c) emits an alpha particle. Solution (a) Z increases by 1, A is unchanged. (b) Z decreases by 1, A is unchanged. (c) Z decreases by 2, A decreases by 4. 5
  • 6. Problem 4 How many successive α-decays occur in the decay of the thorium isotope Th into the lead 228 90 isotope Pb ? 212 82 Solution Each α-decay means a reduction of 2 in atomic number and 4 in mass number. Here Z decreases by 8 and A by 16, which means that 4 α- particles are emitted. HALF LIFE, t 1/2 = ln 2 / λ The time required for half of the atoms in any given quantity of a radioactive isotope to decay is the half-life of that isotope. Each particular isotope has its own half-life. For example, the half-life of 238U is 4.5 billion years. That is, in 4.5 billion years, half of the 238U on Earth will have decayed into other elements. In another 4.5 billion years, half of the remaining 238U will have decayed. One fourth of the original material will remain on Earth after 9 billion years. The half-life of 14C is 5730 years, thus it is useful for dating archaeological material. 6
  • 7. Nuclear half-lives range from tiny fractions of a second to many, many times the age of the universe. (http://www.lbl.gov/abc/Basic.html). Problem 5 Tritium is the hydrogen isotope H whose nucleus 3 1 contains two neutrons and one proton. Tritium is beta-radioactive and emits an electron. (a) What does tritium become after undergoing beta decay? (b) The half-life of tritium is 12.5 year. How much of the 1-g sample will remain undecayed after 25 years? Solution (a) in the beta decay of nucleus one of the neutrons become a proton. Since the atomic number A= 2 corresponds to helium, the beta decay of tritium is 3 1 H → 23 He + e − and the new atom is 3 2 He . (b) 25 years = 2 x t1/2 , and so ½ x ½ x 1 g = ¼ g of tritium remains undecayed. 7
  • 8. Problem 6 The half-life of the sodium isotope Na against 24 11 beta decay is 15 h. how long does it take for ⅞th of a sample of this isotope to decay? After ⅞th has decayed, ⅛th is left,and 1 1 1 1 = × × which is 3 half-lives. Hence the answer 8 2 2 2 is (3)(15) h = 45 h. Problem 7 The carbon isotope carbon-14 ( C ) called 14 6 radiocarbon is beta-radioactive with a half-life of 5600 years. Radiocarbon is produced in the earth’s atmosphere by the action of cosmic rays on nitrogen atoms, and the carbon dioxide of the atmosphere contains a small proportion of radiocarbon as a result. All plants and animals therefore contain a certain amount of radiocarbon along the stable isotope of C . When 12 6 a living thing dies, it stops taking in radiocarbon and the radiocarbon it already contains decays steadily. By measuring the ratio between C and 12 6 C of the remains of an animal or plant and 14 6 8
  • 9. comparing it with the ratio of these isotopes in living organisms, the time that has passed since the demise of the animal or the plant can be found. (a) How old is a piece of wood from an ancient dwelling if its relative radiocarbon content is ¼th that of a modern specimen? (b) If it is 1/16th that of a modern specimen? Solution (a) since ¼ = ½ x ½ , the specimen is two half- lives old, which is 11,200 yr old. (b) since 116 = ½ x ½ x ½ x ½ , the specimen is 4 half-lives old, which is 22,400 yrs old. 9
  • 10. Problem 8 The rate at which a sample of radioactive substance decays is called its activity. A unit of activity is the curie(Ci) where 1 curie =3.7 * 1010 decays/s. if a luminous watch dial contains 5 µCi of the radium isotope Ra , how many 226 88 decays per second occur in it? (This isotope is – alpha-radioactive). Solution Since 1 µCi is 10-6 Ci, the activity of the watch dial is ( 3.7 ×10 decays / s ) (5 x 10-6 Ci) = 1.85 x 105 decays/s. 10 Ci 10