Ch. 18 Nuclear Chemistry

 So
far we’ve studied chemical
reactions where only electrons have
 Chemical properties are determined by
› Nucleus was not primarily important in these
reactions, as it did not undergo any changes.
› Identities remained the same in chemical
reactions because protons remained the
 This
is no longer true in nuclear
 Nucleus
is extremely small, dense, and
contains a huge amount of energy.
› Millions of times more E than chemical
 Nucleus
= neutrons + protons
› Made of even smaller parts, such as quarks.
where A = mass # & Z = charge/#
of protons.
› Isotopes = atoms of the same element with
different #’s of neutrons (protons stay the
Radioactive Decay &
Nuclear Stability
 Already
discussed these in the packet
you completed.
 Additional note: decay types can be
broken into two categories: those that
change mass # and those that don’t.
 Changes mass #: alpha emission.
› Giving off a He atom decreases mass.
 Example: 23892U  42He + 23490Th
› This is a type of spontaneous fission –
splitting a heavy nuclide into 2 lighter
Do not change mass #: particle
emitted/used has no mass (mass # = 0).
Beta emission: 13153I  0-1e + 13154Xe
Gamma ray emission:
238 U  4 He + 234 Th + 0 
Positron emission:
22 Na  0 e + 22 Ne
Electron capture:
201 Hg + 0 e  201 Au + 0 
 Radiation
emitted has different levels
of penetration.
 The more penetrating the emission,
the more dangerous.
 The order is as follows:
alpha < beta/positron < gamma ray
 Therefore, alpha particles are least
penetrating and gamma rays are by
far most penetrating.
Which of the following statements is true
about beta particles?
are electrons with a mass number of
0 and a charge of -1.
b)They have a mass number of 0, a charge
of -1, and are less penetrating than α
c)They are electrons with a charge of +1
and are less penetrating than α particles.
d)They have a mass number of 0 and a
charge of +1.
When 22688Ra decays, it emits 2 α particles,
followed by a β particle, followed by an α
particle. The resulting nucleus is:
a) 21283Bi
b) 22286Rn
c) 21482Pb
d) 21483Bi
of 3 α particles, so subtract 12 from mass #
and 6 from atomic #: 21482Pb
β particle means get rid of a neutron and add a
proton (and an electron): 21483Bi
The formation of 23090Th from 23492U occurs by:
b)α decay.
d)Positron decay.
An atom of 23892U undergoes radioactive
decay by α emission. What is the product
 If
a nucleus is unstable it will
undergo radioactive decay to
become stable.
 Can be tricky to determine if a
nuclide is stable and how it will
decay, but several
generalizations have been.
› Note: nuclide = a specific nucleus of an
isotope or atom.
 Of
2000 known nuclides, only 279
are stable.
› Tin has the greatest number of stable
isotopes at 10.
 Ratio
of neutrons: protons
determine stability.
• What is the stable
ratio of n:p+ at the
lower end of the
• What is the stable
ratio of n:p+ at the
upper end of the
• Based on this
information, what
can you conclude
about the ratio of
n:p+ in stable
• This can be found
on pg. 842 in your
 Developed
by plotting # of neutrons
vs. # of protons of known, stable
 Low end of the belt shows a stable
ratio of about 1n:1p+.
 As the belt gets higher (more protons),
the stable ratio begins to increase to
about 1.5n:1p+.
 For isotopes with less than 84 p+, the
ratio of n:p+ is a good way to predict
 For
light nuclides (<20 p+), 1:1 ratio of
n:p+ are stable.
 For heavier nuclides (20 to 83 p+) ratio
increases (to ~1.5:1).
› Why?
› More neutrons needed to stabilize
repulsive force of more protons.
 All
nuclides with 84 or more p+ are
unstable (because they’re so big).
› Alpha decay occurs- giving off a He
atom lessens both mass and atomic #.
 Even
#’s of n & p+ are more stable
than odd #’s.
Pg. 843 in textbook
 Magic
Numbers: certain #’s of n or p+
give especially stable nuclides.
 2,8,20,28,50,82,126
› A nuclide with this number of n or p+ would
be very stable.
› If there is a magic number of n & p+, this is
called a double magic number (usually
seen in heavier nuclides were extra stability
is needed).
The stability of magic numbers is similar to
atoms being stable with certain #s of e› 2(He), 8(Ne), 18(Ar), 36(Kr), 54(Xe), 86(Rn)
1) Is the isotope Ne-18 stable?
› 10p+ and 8n  0.8:1
› Ratio is <1:1, so it is unstable!
› Undergoes decay to either increase
#n & decrease #p+
2) Is the isotope 126C stable?
› Yes! Ratio = 1:1 (also even # n & p+)
3) Is the isotope sodium-25 stable?
› Ratio of n:p+ = 14:11  1.3:1
› Not stable! How will it become stable?
› Decrease #n  beta emission
Pg. 869 #3, 4, 12, 13, 20
The Kinetics of Radioactive Decay
Rate of decay = - change in #of nuclides
change in time
› Negative sign = number decreasing
› Tough to predict when a certain atom
will decay, but if a large sample is
examined trends can be seen.
Trends indicate that radioactive decay
follows first-order kinetics.
Thus first-order formulas are used!
Two formulas are used to solve calculations
involving decay and half-life:
 ln[A]t – ln[A]0 = -kt
 t1/2 = ln2
• Usually takes 10 half lives for a radioactive
sample to be ‘safe’.
• Half-lives can be seconds or years!
Formulas above are used when multiples of
half lives are not considered.
Example #1: I-131 is used to treat thyroid
cancer. It has a half life of 8 days. How
long would it take for a sample to decay
to 25% of the initial amount?
Each half-life cycle decreases the initial
amount by half: after 8 days, one halflife, 50% remains. After another 8 days,
25% remains. Thus it would take 16 days.
Every half-life cycle will follow the
following order of percentages of
radioactive isotopes that remain :
100%, 50%, 25%, 12.5%, 6.25%, 3.12%, etc.
 If a question asks about half-life and
involves one of these ‘easy’
percentages, you can simply count how
many half-life cycles have occurred.
 However, not all are this simple!
Example #2: What is the half-life of a
radioactive isotope (radioisotope) that
takes 15min to decay to 90% of its
original activity?
90% is not a multiple of half-life cycles, so
we need to use the previously
mentioned equations to calculate this.
(1) Use ln[A]t – ln[A]0 = -kt to solve for the rate
constant, k.
• Often times specific amounts/concentrations
will not be given! Usually given as
percentages of original sample left over. Just
assume 100 as the original ([A]0) and use the
percent asked about as [A]t.
(2) Then use t1/2 = ln2 & value of k from above
to solve for the half life (units will vary
depending on what is given in the problem).
Example #2: What is the half-life of a
radioactive isotope (radioisotope) that takes
15min to decay to 90% of its original activity?
ln(90) – ln(100) = -k(15min)  k = 0.00702/min
 t1/2 = ln2/0.00702min-1  t1/2 = 98.7min
If both the half-life of a radioactive
isotope is given and the amount of
radioactive substance remaining, the
amount of time it took for this substance
to decay to this point can be
 First, use t1/2 formula to find k.
 Then, use other formula to solve for t.
 This is how carbon dating (uses
radioactive C-14 isotope) is used to
determine ages of objects!
If a wooden tool is discovered, and its C14 activity has decreased to 65% of its
original amount, how old is the tool? The
half-life of C-14 is 5,730 years.
5,730yr = ln2/k  k = 1.21 x 10-4/year
 ln65 – ln100 = -(1.21 x 10-4/year)t
t = 3,600 years
During nuclear reactions and nuclear
decay, energy is given off.
› Gamma rays, x-rays, heat, light, and kinetic E
Why does E always accompany these
› Small amount of matter is turned into E.
› Law of conservation of matter is not followed
during nuclear reactions!
Einstein’s equation is used to perform
calculations involving the mass-energy
change in nuclear reactions: E = mc2.
› E = energy released
› m = mass converted into energy (units need
to be in kg in order to get J as unit of E)
› c = speed of light = 3.00 x 108 m/s
Note that the amount of matter turned
into E in a nuclear reaction is small, but is
amplified by the speed of light to
produce a lot of E!
When one mole of uranium-238 decays
into thorium-234, 5 x 10-6kg of matter is
changed into energy. How much energy
is released during this reaction?
E = (5 x 10-6kg)(3.00 x 108m/s)2
E = 5 x 1011 J
Pg. 870 # 21, 24, 35