Ch. 18 Nuclear Chemistry
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Transcript Ch. 18 Nuclear Chemistry
So
far we’ve studied chemical
reactions where only electrons have
changed.
Chemical properties are determined by
electrons!
› 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
same.
This
is no longer true in nuclear
reactions!
Nucleus
is extremely small, dense, and
contains a huge amount of energy.
› Millions of times more E than chemical
reaction.
Nucleus
= neutrons + protons
› Made of even smaller parts, such as quarks.
AZX
where A = mass # & Z = charge/#
of protons.
› Isotopes = atoms of the same element with
different #’s of neutrons (protons stay the
same).
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
nuclides.
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
92
2
90
0
Positron emission:
22 Na 0 e + 22 Ne
11
1
10
Electron capture:
201 Hg + 0 e 201 Au + 0
80
-1
79
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?
a)They
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
particles.
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
Total
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:
a)Electron
capture.
b)α decay.
c)βdecay.
d)Positron decay.
An atom of 23892U undergoes radioactive
decay by α emission. What is the product
nuclide?
a)23090Th.
b)23490Th.
c)23092U.
d)23091Pa.
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
belt?
• What is the stable
ratio of n:p+ at the
upper end of the
belt?
• Based on this
information, what
can you conclude
about the ratio of
n:p+ in stable
nuclides?
• This can be found
on pg. 842 in your
textbook.
Developed
by plotting # of neutrons
vs. # of protons of known, stable
isotopes.
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
stability.
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
k
• 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.
Steps:
(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
k
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
calculated.
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
reactions?
› 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