Transcript Nuclear

Review
Atomic number (Z) = number of protons in nucleus
Mass number (A) = number of protons + number of neutrons
= atomic number (Z) + number of neutrons
A
Z
Mass Number
Atomic Number
proton
1p
1
or 11H
neutron
1n
0
X
Element Symbol
electron
0e
-1
or
0b
-1
positron
0e
+1
or
0b
+1
a particle
4He
2
or
A
1
1
0
0
4
Z
1
0
-1
+1
2
2
4a
2
Nuclear Equations
• The total number of protons and neutrons before a
nuclear reaction must be the same as the total
number of nucleons after reaction.
• There are three main types of radiation which we
consider:
– a-Radiation is the loss of 42He from the nucleus,
– b-Radiation is the loss of an electron from the
nucleus,
– -Radiation is the loss of high-energy photon from
the nucleus.
• Symbols for other particles are given
1
1
below: Proton
1H or 1P
Copyright © Cengage
Learning. All rights
Neutron
1
0
Electron
0
-1
Positron
0
1
Gamma photon
0
0
n
e or -01β
e or 01β
20 | 4
γ
• In nuclear chemistry to ensure conservation of nucleons
we write all particles with their atomic and mass
numbers: 42He and 42a represent a-radiation.
6
• There are six common types of
radioactive decay.
1.
Alpha emission
Emission of an alpha particle from an unstable
nucleus.
Copyright © Cengage
Learning. All rights
20 | 7
2.
Beta emission
Emission of a beta particle from an unstable
nucleus. Beta emission is equivalent to a
neutron converting to a proton.
Copyright © Cengage
Learning. All rights
20 | 8
3.
Positron emission
Emission of a positron particle from an
unstable nucleus. Positron emission is
equivalent to a proton converting to a neutron.
Copyright © Cengage
Learning. All rights
20 | 9
4.
Electron capture
The decay of an unstable nucleus by capture
of an electron from an inner orbital of the
atom. Electron capture is equivalent to a
proton converting to a neutron.
Copyright © Cengage
Learning. All rights
20 | 10
5.
Gamma emission
Emission from an excited nucleus of a gamma
photon, corresponding to radiation with a
wavelength of approximately 10-12 m.
Technetium-99m is an example of a
metastable nucleus; it is in an excited state
and has a lifetime of ≥ 10-9 s.
Copyright © Cengage
Learning. All rights
20 | 11
6.
Spontaneous fission
The spontaneous decay of an unstable
nucleus in which a heavy nucleus of mass
number greater than 89 splits into lighter
nuclei and energy is released.
Copyright © Cengage
Learning. All rights
20 | 12
•Nucleons can undergo decay:
1 n
0
1
+ 
p
0
 11p+ + 0-1e- (b-emission)
1
0 e+ (positron or b+-emission)
n
+
0
1
• A positron is a particle with the same mass as
an electron but a positive charge.
0
- + 0 e+  20  (positron annihilation)
e
-1
1
0
1 p+
1
+ 0-1e-  10n (electron capture)
Types of Decay:
Alpha: (slowest moving; highest mass
4
2
U  He
238
92
“parent”
211Bi 
83
+
234
90
Th
“daughter”
Molecular view of the nuclear
equation for the decay of uranium238
Beta: (e-)
131
53
I 
234Th 
90
n
1
0
0
-1
e 
e  11p
0
-1
131
54

Xe
Gamma: (quite often occurs in conjunction with other decay processes)
60Co*  60Co  
27
27
(extremely penetrating)
Molecular view of the nuclear
equation for the decay of
technetium-99 (gamma emission)
Kelter, Mosher and Scott, Chemistry: The Practical Science, 1/e.
Positron emission:




1 e or 0b or b  

0
1





1 p 1n  0e 
1
0
1 
95Tc  95Mo  0e
1
42
43
PET Scan:
0
- + 0 e+  20  (positron annihilation)
e
-1
1
0
Molecular view of the nuclear
equation for the decay of
technetium-95 (positron emission)
Figure 20.17: A patient undergoing
a PET scan of the brain
Alexander Tsiara/ Photo Researchers, Inc.
Figure 20.16: PET scans of normal
and schizophrenic patients
Wellcome/ Photo Researchers, Inc.
electron capture:
1 p  0e  1n  x - ray
1
-1
0
40K  0e  40Ar  x - ray
19
-1
18
Molecular view of the nuclear
equation for the decay of
potassium-40 (electron capture)
Patterns of Nuclear Stability
Neutron-to-Proton Ratio
• The proton has high mass and high charge.
• Therefore the proton-proton repulsion is large.
• In the nucleus the protons are very close to each other.
• The cohesive forces in the nucleus are called strong
nuclear forces. Neutrons are involved with the strong
nuclear force.
• As more protons are added (the nucleus gets heavier)
the proton-proton repulsion gets larger.
26
n/p too large
beta decay
X
Y
n/p too small
positron decay or electron capture
27
Neutron-to-Proton Ratio
Neutron-to-Proton Ratio
•The heavier the nucleus, the more
neutrons are required for stability.
•The belt of stability deviates from a 1:1
neutron to proton ratio for high atomic
mass.
•At Bi (83 protons) the belt of stability ends
and all nuclei are unstable.
Stable
–Nuclei above the belt of stability
undergo b-emission. An electron is lost
and the number of neutrons decreases,
the number of protons increases.
• For stable nuclides with Z ≤ 20, the ratio of
neutrons to protons is between 1 and 1.1.
• For stable nuclides with Z > 20, the ratio of
neutrons to protons increases to about
1.5. This is believed to be due to the
increasing repulsion between protons,
which requires more neutrons to increase
the strong nuclear force.
Copyright © Cengage
Learning. All rights
20 | 29
Nuclear Stability and Radioactive Decay
Beta decay
14C
6
+ 0b
14N
40K
19
7
40Ca
20
Decrease # of neutrons by 1
-1
+ 0b
Increase # of protons by 1
-1
1n
0
1p
+ 0b
1
-1
Positron decay
+ 0b
11C
11B
38K
38Ar
6
19
5
18
Increase # of neutrons by 1
+1
+ 0b
Decrease # of protons by 1
+1
1p
1
1n
+ 0b
0
+1
30
Nuclear Stability and Radioactive Decay
Electron capture decay
37Ar
+ 0e
37Cl
55Fe
+ 0e
55Mn
18
26
Increase number of neutrons by 1
17
-1
Decrease number of protons by 1
25
-1
1p
1
+ 0e
-1
1n
0
Alpha decay
Decrease number of neutrons by 2
212Po
84
4He
2
+ 208Pb
82
Decrease number of protons by 2
Spontaneous fission
252Cf
98
2125In + 21n
49
0
31
Nuclear binding energy is the energy required to break up a nucleus
into its component protons and neutrons.
Nuclear binding energy + 19F
9
91p +
1
101n
0
DE = (Dm)c2
9 x (p mass) + 10 x (n mass) = 19.15708 amu
Dm= 18.9984 amu – 19.15708 amu
Dm = -0.1587 amu
DE = -0.1587 amu x (3.00 x 108 m/s)2
= -1.43 x 1016 amu m2/s2
Using conversion factors:
1 kg = 6.022 x 1026 amu
1 J = kg m2/s2
DE = -2.37 x 10-11J
32
DE = (-2.37 x 10-11J) x (6.022 x 1023/mol)
DE = -1.43 x 1013J/mol
DE = -1.43 x 1010kJ/mol
Nuclear binding energy = 1.43 x 1010kJ/mol
binding energy
binding energy per nucleon =
number of nucleons
=
2.37 x 10-11 J
19 nucleons
= 1.25 x 10-12 J/nucleon
33
Nuclear binding energy per nucleon vs mass numbe
nuclear binding energy
nucleon
nuclear stability
34
Patterns of Nuclear Stability
Neutron-to-Proton Ratio
– Nuclei below the belt of
stability undergo b+-emission
or electron capture. This
results in the number of
neutrons increasing and the
number of protons
decreasing.
Nuclei with atomic numbers
greater than 83 usually
undergo a-emission. The
number of protons and
neutrons decreases (in
steps of 2).
Patterns of Nuclear Stability
Radioactive Series
For 238U, the first decay is to
234Th (a-decay). The 234Th
undergoes b-emission to 234Pa
and 234U. 234U undergoes adecay (several times) to 230Th,
226Ra, 222Rn, 218Po, and 214Pb.
214Pb undergoes b-emission
(twice) via 214Bi to 214Po which
undergoes a-decay to 210Pb.
The 210Pb undergoes b-emission
to 210Bi and 210Po which decays
(a) to the stable 206Pb.
37
Radiocarbon Dating
14N
7
+ 1n
14C
6
14C
6
0
14N
7
+
+ 1H
1
0b
t½ = 5730 years
-1
Uranium-238 Dating
238U
92
206Pb
82
+ 8 4a + 6 0b
2
-1
t½ = 4.51 x 109 years
38
Kelter, Mosher and Scott, Chemistry: The Practical Science, 1/e. Copyright © 2008 by Houghton Mifflin Company. Reprinted with permission.
Figure 20.19: Representation of a
chain reaction of nuclear fissions
Kelter, Mosher and Scott, Chemistry: The Practical Science, 1/e. Copyright © 2008 by Houghton Mifflin Company. Reprinted with permission.
Figure 20.20:
An atomic
bomb
Figure 20.21: Light-water nuclear
reactor
Molecular views of nuclear fusion
reactions of deuterons with
deuterium or tritium
•
Copyright © Cengage
Learning. All rights
20 | 44
• Nuclear Bombardment Reactions
• Nuclear bombardment reactions are not
spontaneous. They involve the collision of
a nucleus with another particle.
• Transmutation is the change of one
element into another by bombarding the
nucleus of the element with nuclear
particles or nuclei.
Copyright © Cengage
Learning. All rights
20 | 45
• When Rutherford allowed alpha particles
to collide with nitrogen nuclei, he found
that a proton was ejected and oxygen was
formed.
Copyright © Cengage
Learning. All rights
20 | 46
• Sodium-22 is made by the
bombardment of magnesium-24 (the
most abundant isotope of magnesium)
with deuterons. An alpha particle is
the other product.
Reaction :
24
12
Mg  H 
2
1
Abbreviated notation :
Copyright © Cengage
Learning. All rights
20 | 47
24
12
22
11
Na  He
4
2
Mgd, a  Na
22
11
• Half-life is the time it takes for one-half of
the nuclei in a sample to decay.
• Half-life is related to the decay constant by
the following equation:
t1
Copyright © Cengage
Learning. All rights
2
0.693

k
20 | 48
•
Copyright © Cengage
Learning. All rights
20 | 49
• Thallium-201 is used in the diagnosis
of heart disease. This isotope decays
by electron capture; the decay
constant is 2.63 × 10-6/s. What is the
half-life of thallium-201 in days?
Copyright © Cengage
Learning. All rights
20 | 50
t1
2
0.693
t1 
2
k
0.693
t1 
2
2.63  10 -6
s
1 min
1h
1 day
5
 2.63  10 s 


60 s 60 min 24 h
t 1  3.05 days
2
Copyright © Cengage
Learning. All rights
20 | 51
• The rate constant is related to the fraction
of nuclei remaining by the following
equation:
N 
ln t   - kt
 N0 
N0 is the original number of nuclei.
Nt is the number of nuclei at time t.
Nt
is the fractionof nuclei remaining at time t.
N0
Nt  4.745 x 10 nuclei
15
Copyright © Cengage
Learning. All rights
20 | 52
• A 0.500-g sample of iodine-131 is
obtained by a hospital. How much will
remain after a period of one week?
The half-life of this isotope is 8.07
days.
Copyright © Cengage
Learning. All rights
20 | 53
• First, we find the value of k.
0.693
k
t1
2
0.693
k
1 w eek
8.07 days 
7 day
0.601
k
week
Copyright © Cengage
Learning. All rights
20 | 54
Next, we find the fraction of nuclei remaining.
 Nt 
  - kt
ln
 N0 
 Nt 
0.601
ln     1 w eek
w eek
 N0 
 Nt 
  - 0.601
ln
 N0 
 Nt 
   0.548
 N0 
54.8% of nuclei remain.
Copyright © Cengage
Learning. All rights
20 | 55
• Radioactive Dating
• Because the rate of radioactive decay is
constant, this rate can serve as a sort of clock
for dating objects.
• Carbon-14 is part of all living material. While a
plant or animal is living, the fraction of carbon-14
in it remains constant due to exchange with the
atmosphere. Once dead, the fraction of carbon14 and, therefore, the rate of decay decrease. In
this way, the fraction of carbon-14 present in the
remains becomes a clock measuring the time
since the plant’s or animal’s death.
Copyright © Cengage
Learning. All rights
20 | 56
• The half-life of carbon-14 is 5730 years.
Living organisms have a carbon-14 decay
rate of 15.3 disintegrations per minute per
gram of total carbon.
• The ratio of disintegrations at time t to time
0 is equal to the ratio of nuclei at time t to
time 0.
Copyright © Cengage
Learning. All rights
20 | 57
• A sample of wheat recovered from a
cave was analyzed and gave 12.8
disintegrations of carbon-14 per
minute per gram of carbon. What is
the age of the grain?
• Carbon from living material decays at
a rate of 15.3 disintegrations per
minute per gram of carbon. The halflife of carbon-14 is 5730 years.
Copyright © Cengage
Learning. All rights
20 | 58
• Ratet = 12.8 disintegrations/min/g
• Rate0 = 15.3 disintegrations/min/g
• t1/2 = 5730 y
 N t  rate t 12.8 disintegrations/min/ g

 
 0.8366

 N 0  rate 0 15.3 disintegrations/min/ g
 Nt 
 Nt 


ln 
ln 
N0 
N0 
ln 0.8366




t
t1
 0.693 
 0.693 


  
2
 k 
 5730 y 
t  1.48  10 y
3
Copyright © Cengage
Learning. All rights
20 | 59
• Energy of Nuclear Reactions
• Nuclear reactions involve changes of
energy on a much larger scale than occur
in chemical reactions. This energy is used
in nuclear power reactors and to provide
the energy for nuclear weapons.
Copyright © Cengage
Learning. All rights
20 | 60
• Mass–Energy Calculations
• When nuclei decay, they form products of
lower energy. The change of energy is
related to changes of mass, according to
the equation derived by Einstein, E = mc2.
Copyright © Cengage
Learning. All rights
20 | 61
• Nuclear Binding Energy
• The equivalence of mass and energy
explains the mass defect—that is, the
difference between the total mass of the
nucleons that make up an atom and the
mass of the atom. The difference in mass
is the energy holding the nucleus together.
• The binding energy of a nucleus is the
Copyright © Cengage
20 | 62
Learning.
All rights needed to break a nucleus into its
energy
•
The maximum binding
energy per nucleon
occurs for nuclides
with mass numbers
near 50
Copyright © Cengage
Learning. All rights
20 | 63
Energy Changes in Nuclear Reactions
 23490Th + 42He
for 1 mol of the masses are
238.0003 g  233.9942 g + 4.015 g.
The change in mass during reaction is
233.9942 g + 4.015 g - 238.0003 g = -0.0046 g.
The process is exothermic because the system has
lost mass.
To calculate the energy change per mole of 23892U:
238
–
–
–
–
92U
DE  D mc 2  c 2 Dm


2
 1 kg 
8
 3.00  10 m/s  0.0046 g 

 1000 g 
kg - m 2
11
 4.1  10
 4.1  1011 J
s2
DE  D mc 2  c 2 Dm


2
 1 kg 
8
 3.00  10 m/s  0.0046 g 

 1000 g 
kg - m 2
11
 4.1  10
 4.1  1011 J
s2
Radioisotopes in Medicine
Research production of 99Mo
98Mo
42
+ 1n
Bone Scan with
99mTc
99Mo
42
0
Commercial production of 99Mo
235U
92
99Mo
42
99mTc
43
+ 1n
0
99Mo
+ other fission products
+ 0b
t½ = 66 hours
+ -ray
t½ = 6 hours
42
99mTc
43
99Tc
43
-1
66
Chemistry In Action: Food Irradiation
67