Ch. 24--Nuclear Chemistry
“It’s all about the nucleus!”
Radioactivity
Becquerel (1896)--Found that some of his
film was exposed inside its envelope
after uranium salts were placed near it.
Something penetrated the paper that
could expose the film!
Radioactivity
Marie and Pierre Curie, students of
Becquerel, discovered that invisible rays
given off by the Uranium were
responsible.
Marie called the process of giving off
these rays radioactivity.
Radioactivity
Marie, one of science’s most famous
female scientists, went on to “discover”
two radioactive elements: Polonium
(named for her native Poland) and
radium.
She is the only person to win Nobel
Prizes in two different sciences.
Types of Radiation
• Three types: alpha, beta, & gamma
Types of Radiation
• Alpha ( ) radiation
– consists of helium nuclei, emitted by a
radioactive source.
– Alpha particles have 2 protons, 2 neutrons,
and thus a +2 charge.
– Written as

4
2
He or simply

Types of Radiation
) radiation
• Beta (
--consists of fast electrons, made as
neutrons decompose:
1
0
n  p +
1
1
0
-1
e
--Beta particles are simply electrons, they
have a mass number of zero, and are
given an atomic number of -1.

--Written as
0
-1
e
or simply

Types of Radiation
• Gamma (
) radiation
– Is energy that is released, there is no particle.
– Gamma rays are emitted at the same
time as alpha or beta, or sometimes just
as a energy release as an excited nucleus
comes to its ground state:
Reasons for radioactivity
Some isotopes are radioactive; we
call these radioisotopes.
Radioisotopes have unstable
nuclei. This lack of stability leads
to a “shuffling” of the nucleus in
search of a more stable nucleus.
Why is nucleus unstable?
1) Too many n0/too few p+
2) Too few n0/too many p+
3) Nucleus too heavy! (too many
n0 and p+)
Decay Reactions
• An unstable isotope’s nucleus breaks
apart, giving off a small particle and
forming a new larger particle.
• Original isotope = Mother
• New isotope = Daughter
Particles
Alpha
Beta
Positron

neutron

Deutron



 orHe
4
2
 or
0
-1
0
+1
e
1
0
n
2
1
H
Proton
1
1
H
e
Not a particle:

Gamma 
Writing Nuclear Equations
• Like any equation, both sides must be
equal!
• The arrow is our equal sign
• The mass numbers on each side must
be equal
• The atomic numbers on each side must
be equal
Writing nuclear equations
If you’re told the type of reaction it is
supposed to be, that tells you what
particles are to be involved!
Alpha Decay
In an alpha decay, the mother breaks into
two particles: an alpha particle, and the
daughter.
Write the mother on the left, draw the
arrow, write the alpha particle formula
on the right. Balance, and use P. Table
to ID the new daughter.
Alpha Decay
Write the equation showing the alpha decay of
a uranium-233 isotope.
233
92
U 


4
2
He 

229
90
Th
Beta Decay
In an beta decay, the mother breaks into
two particles: a beta particle, and the
daughter.
Write the mother on the left, draw the
arrow, write the beta particle formula on
the right. Balance, and use P. Table to
ID the new daughter.
Beta Decay
Write the equation showing the beta
decay of a iridium-192 isotope.
192
77
Ir



0
-1
e


192
78
Pt
Other Possibilities
• “Electron capture” This reaction has an
electron () as a reactant. The isotope
on the left is gaining the electron
instead of shedding it.
Show electron capture for La-137
137
57
La +
0
-1
e

137
56
Ba
Other Possibilities
• Positron emission: “To emit” something
means “to give off.” Here, a mother
0
gives off a positron (+1 e) and a daughter.
Show the positron emission reaction of zinc-65:
65
30

Zn

0
+1
e +
65
29
Cu
Summing Up
If we know our vocab, it’s not too bad!
“Decay” or “emission” reactions will have
one reactant, two products.
“Capture” reactions will have two
reactants, one product.
Series reactions
• In order to become stable, a
radioisotope will often have to undergo
a number of successive decays.
Series Reactions
Show how Fr-222 undergoes successive beta,
alpha, alpha, and alpha decays.
222
87
222
88



218
86
Fr
 e +
Ra
 He +
218
86
Rn
 He +
214
84
Po  He +
210
82
214
84
0
-1
4
2
4
2
4
2
222
88
Ra
Rn
Po
Pb
Types of nuclear reactions
1. Natural Decay: An atom decays, giving
off a beta, alpha, or positron.
2. Induced decay: An atom’s nucleus is
“bombarded” with a proton, neutron, or
alpha particle, which destabilizes the
nucleus and causes a nuclear reaction.
Types of nuclear reactions
3. Electron capture: An electron converts
a proton to a neutron
4. Fission Reaction: A large nucleus
cleaves into two smaller (yet larger
than alpha!) nuclei
5. Fusion: Two smaller nuclei combine to
form a larger nucleus.
Half-life
• The half-life of a radioisotope is the
amount of time it takes for 1/2 of a
sample to decay.
• This amount of time is unique for each
radioisotope.
• The time does not change based on the
amount of material present!
Half-life
A substance with t1/2 = 22 days
Time:
0 days
22 days
44 days 66 days
# 1/2lives
Mass:
0
1
2
3
1000 kg
500 kg
250 kg
125 kg
5 mg
2.5 mg
1.25 mg 0.625 mg
100% (1)
50% (1/2)
25% (1/4) 12.5% (1/8)
Sample 1
Mass:
Sample 2
% orig.
sample
Half-life
Things not to forget when working half-life problems:
• The amount of time for a half-life never changes; the
amount of mass that is decayed in that time does change!
• The initial time is always “0.” Zero time has passed, zero
half-lifes have passed.
•The fraction of the original isotope remaining can be
figured out several ways.
1 
  
2 
# half -lives
amount remaining
is one.
= (original amount)
Half-life
Another way to do half-life problems is to make a 3column chart, that features number of half-lives, the
amount of time that’s passed, and the amount of
sample that remains.
Let’s consider problem 14 from our HW: Actinium-226
has a half-life of 29 hours. If 100 mg of
actinium-226 disintegrates over a period of 58
hours, how many mg of actinium-226 will
remain?
# of half-lives Amt of time
0
0
Amt of 226Ac
1000 mg
# of half-lives Amt of time
Amt of 226Ac
0
0
1000 mg
1
29 hrs
500 mg
# of half-lives Amt of time
Amt of 226Ac
0
0
1000 mg
1
29 hrs
500 mg
2
58 hrs
250 mg
We could also solve this problem by
recognizing that 2 half lives go by in the
58 hours ( 58 hrs  1 half life = 2 half - lives ) and
29 hrs
either using our half-life curve to see
that 25% of our original sample is left, or
by plugging into the equation
amount remaining
= (original amount)
1 # half -lives
  
2 