chapter 19 radioactivity and nuclear chemistry 2 goals types of radioactivity identify radioactive...
TRANSCRIPT
Chapter 19
Radioactivity and Nuclear Chemistry
2
GOALS
Types of radioactivityIdentify radioactive nuclidesNuclear equationsBinding energy; per nucleon; unitsKinetics of radioactive decay
3
Facts About the Atomic Nucleus
• Every atom of an element has the same number of protons (+ve)atomic number (Z)
• Atoms of the same elements can have different numbers of neutrons (no charge)
• Isotopes: atoms of the same element, having same atomic number, Z, but different mass number, A (diff no. of neutrons).
• Isotopes are identified by their mass number (A)mass number = number of protons + neutrons
4
Facts About the Atomic Nucleus
• mass number = number of protons + neutrons
neutrons = mass number – number of protons
• The nucleus of an isotope is called a nuclide
• Each nuclide is identified by a symbol
X Element AZ
number massnumber atomic
Nuclide Nuclide SymbolsSymbols
• Boron-10 (Boron-10 (101055B) has 5 p and 5 n B) has 5 p and 5 n
• Boron-11 (Boron-11 (111155B) has 5 p and 6 nB) has 5 p and 6 n
10B
11B
• Oxygen-16 (Oxygen-16 (161688O) has 8 p and 8 n O) has 8 p and 8 n
• Oxygen-17 (Oxygen-17 (171788O) has 8 p and 9 nO) has 8 p and 9 n
• Oxygen-18 (Oxygen-18 (181888O) has 8 p and 10 nO) has 8 p and 10 n
6
The Discovery of Radioactivity• Becquerel discovered that certain minerals were
constantly producing penetrating energy rays he called uranic rays (1896)
Marie Curie discovered 2 new elements (Po, Ra) which also emitted uranic rays.
• Curie changed term uranic rays to radioactivity (present in elements other than uranium).
Some nuclei are unstable; they emit particles and/or electromagnetic radiation spontaneously.
This is radioactivity.
7
Types of Radioactive Rays
• Rutherford discovered there were 3 types of radioactivity;
• 2 additional types were later discovered.
• alpha () & beta () decay, gamma ray (then positron emission, and electron capture.
Another type of radioactivity (nuclear transmutation) results from the bombardment of nuclei (heavy) by neutrons, protons or other nuclei (lighter).
8
Penetrating Ability of Radioactive Rays
0.01 mm 1 mm 100 mmPieces of lead
Particle Symbol Nuclear Symbol
proton p+
neutron n0
electron e-
alpha
beta , -
positron , +9
Important Atomic Symbols
p H 11
11
n10
e01
He α 42
42
e β 01
01
e β 01
01
10
Nuclear Equations• nuclear processes are described using nuclear
equations
• use the symbol of the nuclide to represent the nucleus
• atomic numbers and mass numbers are conserveduse this to predict identity of daughter nuclide if parent and
emitted particle are known
nuclides daughter
He Th U
nuclideparent
42
23490
23892
emitted particle: product
captured particle: reactant
He Th U 42
23490
23892
11
Alpha Emission
• an particle contains 2 protons and 2 neutrons
• most ionizing, but least penetrating
• loss of an alpha particle meansatomic number decreases by 2mass number decreases by 4
Rn He Ra 21886
42
22288
He α 42
42
He Th U 42
23490
23892
12
Beta Emission
• An unstable nucleus emits an electron
• when an atom loses a particle itsatomic number increases by 1mass number remains the same
• in beta decay, a neutron changes into a proton
Pa e Th 23491
01
23490
e β 01
01
If californium-251 decays by successive α, α, β emissions, what nucleus is produced?
a) b) c) d) e)
24395 Am 243
93 Np
24294 Pu
24494 Pu
24792 Pa
14
Gamma Emission• gamma () rays are high energy photons of light
• least ionizing, but most penetrating
• generally occurs after the nucleus undergoes some other type of decay and the remaining particles rearrange
γ00
00
23490
42
23892 Th He U
15
Positron Emission
• The positron has a charge of +1 and negligible massanti-electron
• when an atom loses a positron from the nucleus, itsmass number remains the sameatomic number decreases by 1
• A positron appears to result from a proton changing into a neutron
Ne e Na 2210
01
2211
e β 01
01
n e p 10
01
11
16
Electron Capture
• occurs when an inner orbital electron is pulled into the nucleus
• no particle emission, but atom changessame result as positron emission
• proton combines with the electron to make a neutron
mass number stays the sameatomic number decreases by one
Tc e Ru 9243
01
9244
e01
n e p 10
11
01
17
Summary of Decay Processes
(Table 19.1; pg 871)
Decay Emission At # Mass #
-2 -4 inc
- +1 0 dec
-ray 0 0 -
-1 0 inc
e-capt X-ray -1 0 inc
Z
N
He42
18
Write the nuclear equation for positron emission from K-40
a) Write the nuclide symbols for both the starting radionuclide and the particle
e positron K 04K 01
4019
b) Set up the equation (emitted particles are products; captured particles are reactants)
X e K AZ
01
4019
c) Determine the mass number and atomic number of the missing nuclide (mass and atomic numbers are conserved)
X e K 4018
01
4019
19
4) Determine the element from the atomic number
Ar e K 4018
01
4019
Write the nuclear equation for positron emission from K-40
Q. In a decay series, U-238 emits 8 alpha particles and 6 beta particles. What nuclide is formed?
? He8 e6 U 42
01
23892
Pb X X He8 e6 U 20682
42
01
23892
Mass dec by 32; charge = +6 & -16
20
Write a nuclear equation for each of the following
electron capture by Be-7
positron emission from N-13
beta emission from Ne-24
alpha emission from U-238
Stability Stability of of NucleiNuclei
- stable isotopes stable isotopes fall in a very fall in a very narrow range narrow range called the island called the island of stability.of stability.
22
What Causes Nuclei to Break Down?
• the particles in the nucleus are held together by a very strong attractive force found in the
nucleus called the strong force
acts only over very short distances
• the neutrons play an important role in stabilizing the nucleus, as they add to the strong force, but do not repel each other like the protons do
23
Neutron to Proton (N/Z) Ratio
• the ratio of neutrons : protons is an important measure of the stability of the nucleus
• if the N/Z ratio is too high (neutron rich) – neutrons are converted to protons via decay
• if the N/Z ratio is too low (proton rich) – protons are converted to neutrons via positron emission or electron capture
or via decay – though not as efficient
24
Valley (Island) of Stability (Plot of # Neutrons vs # Protons)
for Z = 1 20 (H - Ca), stable N/Z ≈ 1
for Z = 20 40, stable N/Z approaches 1.25
for Z = 40 80, stable N/Z approaches 1.5
Heavy nuclei: for Z > 83, there are no stable nuclei
low N/Z
25
Determine the kind of radioactive decay that Mg-22 undergoes
• Mg-22Z = 12 (protons)N = 22 – 12 = 10 (neutrons)
• N/Z = 10/12 = 0.83 • from Z = 1 20, stable
nuclei have N/Z ≈ 1
• Mg-22 has low N/Z; it should convert 1
1p into 10n, therefore
it will undergo positron emission or electron capture
26
Determine the kind of radioactive decay that N-18 undergoes
• N-18Z = 7 (protons)N = 18 – 7 = 11 (neutrons)
• N/Z = 11/7 = 1.57 • from Z = 1 20, stable
nuclei have N/Z ≈ 1
27
Q. Which of the following will undergo beta decay?
16O, 20F, 13N
28
Magic Numbers
most stable when N or Z = 2 (He), 8 (O), 20 (Ca), 28 (Ni), 50 (Sn), 82 (Pb)
besides the N/Z ratio, the numbers of protons and neutrons effects stability
most stable nuclei have even numbers of protons and neutrons
only a few have odd numbers of protons and neutrons
if the total number of nucleons adds to a magic number, the nucleus is more stable (compare # electrons in noble gases)
Binding Energy, EBinding Energy, Ebb
--All atoms are a little lighter than they are really
supposed to be..Missing mass: ∆m = mass defect.
-This missing mass is converted to energy, -This missing mass is converted to energy, and released when 1 mole of atoms is and released when 1 mole of atoms is formed from its subatomic particles formed from its subatomic particles (protons + neutrons + electrons). (protons + neutrons + electrons).
--Energy holds the nucleus together. Energy holds the nucleus together.
Calculating Binding Energy, Calculating Binding Energy, EEbb
Eb is the energy required to separate the nucleus of an atom into protons, neutrons, electrons.
For stability, EFor stability, Ebb > electrostatic repulsive forces > electrostatic repulsive forces
between protons.between protons.
In deuterium, In deuterium, 2211HH
2211H H 11
11p + p + 1100nn EEbb = 2.15 = 2.15 10 1088 kJ/mol kJ/mol 22
11HH
EEbb per mol nucleon per mol nucleon = E = Ebb/2 nucleons/2 nucleons
= 1.08 = 1.08 10 1088 kJ/mol nucleons kJ/mol nucleonsAlso, calc EAlso, calc Ebb per nucleon per nucleon ((6.0226.022 10 102323
nucleons) nucleons)
Calculating Binding Energy, Calculating Binding Energy, EEbb
For deuterium, For deuterium, 2211H: H: 22
11H H 1111p + p + 11
00nn
Actual mass of Actual mass of 2211H = 2.01410 g/mol (given or PT)H = 2.01410 g/mol (given or PT)
Mass of proton = Mass of proton = 1.007825 g/mol1.007825 g/mol
Mass of neutron = Mass of neutron = 1.008665 g/mol1.008665 g/mol
Theoretical mass = Theoretical mass = 2.016490 g/mol2.016490 g/mol
Mass defect (‘missing mass’) = Mass defect (‘missing mass’) = 2.0164902.016490 – –
2.014102.01410
= 0.00239 g/mol= 0.00239 g/mol
Calculate Binding Energy, ECalculate Binding Energy, Ebb
Mass defect Mass defect = 0.00239 g/mol = 0.00239 g/mol
= (0.00239 = (0.002391000) kg/mol1000) kg/mol
= 2.39 = 2.39 10 10-6-6 kg/mol kg/mol From Einstein’s equation: From Einstein’s equation:
EEbb = (∆m)c = (∆m)c22 = 2.39 = 2.39 10 10-6-6 kg kg (3.00 × 10 (3.00 × 1088 m/s) m/s)22
= 2.15 ×10= 2.15 ×101111 kg kgmm22/s/s22 (but 1 kg(but 1 kgmm22/s/s22 = 1 J) = 1 J)
= 2.15 = 2.15 10 101111 J/mol J/mol 1000 J = 2.15 1000 J = 2.15 10 1088
kJ/mol kJ/mol Two nucleons for deuterium, Two nucleons for deuterium, 22
11H: H: 1111p + p + 11
00nn
EEbb /mol nucleon = 1.08 /mol nucleon = 1.08 10 1088 kJ/mol nucleons kJ/mol nucleons
Calculating Binding Energy, Calculating Binding Energy, EEbb
For I-127, For I-127, 1271275353I: 53p + 74n (i.e. 127 nucleons)I: 53p + 74n (i.e. 127 nucleons)
Actual mass of Actual mass of 1271275353I = 126.9045 g/mol (given or PT)I = 126.9045 g/mol (given or PT)
53 protons = 5353 protons = 531.007825 g/mol = 53.41473 g/mol1.007825 g/mol = 53.41473 g/mol
74 neutrons = 74 74 neutrons = 74 1.008665 g/mol = 74.64121 g/mol 1.008665 g/mol = 74.64121 g/mol
Theoretical mass defect = Theoretical mass defect = 128.05594 g/mol128.05594 g/mol
Mass defect = (128.05594 -126.9045) g/mol Mass defect = (128.05594 -126.9045) g/mol
= 1.1514 g/mol= 1.1514 g/mol
= = 1.1514 1.1514 10 10-3-3 kg/mol kg/mol
Calculate Binding Energy, ECalculate Binding Energy, Ebb
EEbb = = 1.1514 1.1514 10 10-3-3 kg/mol kg/mol (3.00 × 10 (3.00 × 1088 m/s) m/s)22
= 1.04 ×10= 1.04 ×101414 kg kgmm22/s/s22 (but 1 kg(but 1 kgmm22/s/s22 = 1 J) = 1 J)
= 1.04 ×10= 1.04 ×101414 J/mol J/mol
Also, can express EAlso, can express Ebb in MeV: 1 MeV = 1.602 ×10 in MeV: 1 MeV = 1.602 ×10-13 -13 JJ
EEbb /nucleon = ? MeV /nucleon = ? MeV
EEbb /mol nucleon = 1.04 ×10 /mol nucleon = 1.04 ×101414 J/ (127 nucleons) J/ (127 nucleons)
= 8.19 ×10= 8.19 ×101111 J J
EEbb /nucleon = 8.19×10 /nucleon = 8.19×101111 J J (6.022 ×10 (6.022 ×102323))
= 1.36 ×10= 1.36 ×10-12-12 J J
35
Plot of Eb vs Mass
-the greater the binding energy per nucleon, the more stable the nucleus is
Nuclear FissionNuclear Fission
TThe splitting of a heavy unstable nucleus he splitting of a heavy unstable nucleus of an of an atom into two or more fragments; Pu, U & atom into two or more fragments; Pu, U & Th!Th!
--induced reaction to produce energy!induced reaction to produce energy!n Kr Ba n U 1
09136
14256
10
23592 3
Energy released Energy released 16,800,000,000 16,800,000,000 kJ/mol kJ/mol (235 g Uranium) (235 g Uranium)
Nuclear FusionNuclear Fusion
Free of long-lived radioactive Free of long-lived radioactive wastewaste..
n2 p2 He2 H6 10
11
42
21
More destructive than fission bombs More destructive than fission bombs (WWII)! (WWII)!
More difficult to achieve. Nuclei must More difficult to achieve. Nuclei must travel at v. large KE’s at each othertravel at v. large KE’s at each other..
Light nuclei fuse to generate Light nuclei fuse to generate heavier nuclei (more stable)heavier nuclei (more stable)
38
Kinetics of Radioactive Decay
0
t
0
t
rate
ratelnt
N
Nln k
k
0.693t1/2
Rate = kN It is a first order processN = number of radioactive nuclei
the shorter the half-life, the more nuclei decay
every second (sample is hot!), the higher the rate
1/2t
timelives-half of no.
0NlntNln t k
39
The half life of Pu-236 is 2.86 years. Starting with a 1.35 mg sample of Pu-236, calculate the mass that will remain after 5.00 years
Concept Plan:
Relationships:
mass Pu-236 = 1.35 mg, t = 5.00 yr, t1/2 = 2.86 yr
mass, mg
Given:
Find:
k
693.0t
21 t
m
mln
0
t k
t1/2 k m0, t mt+
k
693.0t
21
1-yr 3224.0yr 86.2
693.0
t
693.0
21
k
40
Starting with a 1.35 mg sample of Pu-236, calculate the mass that will remain after 5.00 years
units are correct, the magnitude makes sense since it is less than ½ the original mass for longer than 1 half-life
Check:
Solve:
t1/2 k m0, t mt+
-1yr 3224.0k
tN
Nln
0
t k
? N t
yr 00.5yr 2423.0t0t
-1
mg 1.35NN ee k
41
An ancient skull gives 4.50 dis/min∙gC. If a living organism gives 15.3 dis/min∙gC, how old is the skull? 14C-t1/2 = 5730 yr
dis = disintegrations
Solve:
Concept Plan:
Relationships:
ratet = 4.50 dis/min∙gC, ratet = 15.3 dis/min∙gC
time, yr
Given:
Find:
k
693.0t
21 t
rate
rateln
0
t k
t1/2 k rate0, ratet t+
1-4 yr 10902.1yr 7305
693.0
t
693.0
693.0t
21
21
k
k
42
An ancient skull gives 4.50 dis/min∙gC. If a living organism gives 15.3 dis/min∙gC, how old is the skull? 14C-t1/2 = 5730 yr
units are correct, the magnitude makes sense since it is less than 2 half-lives
Check:
Solve:
traterate
ln0
t
k
trate
rateln
0
t k
?yr 10901.2
15.3
4.50ln
t1-4-
gC mindis
gC mindis
43
An artifact containing carbon taken from the tomb of a king of ancient Egypt gave 8.1 dpm/gC. How old is the artifact? Carbon from a living organism gives 15.3 dis/min∙gC; 14C-t1/2 = 5730 yr.
dis = disintegrations
1/20
t
t
t0.693 -
rate
rateln
k1/2t
0.693 trate
rateln
0
t k
trate
rateln
693.0 t
02/1 t
t8.1
15.3ln
693.0
5730 t63599.0
693.0
5730
44
• bombardment of one nucleus with another (2H, 4He, 10B, 12C) causing new atoms to be madecan also bombard with neutrons; protons
• reaction done in a particle accelerator linearcyclotron
Tc-97 is made by bombarding Mo-96 with deuterium, releasing a neutron
n Tc H Mo 10
9743
21
9642
Joliot-Curies
Artificial Nuclear Artificial Nuclear ReactionsReactions
Artificial Nuclear ReactionsArtificial Nuclear Reactions
An example of a n, reaction is production of radioactive 31P for use in studies of P uptake in the body.
3115P + 1
0n 3215P +
Reactions using neutrons are called n, reactions because a ray is usually emitted.
Radioisotopes used in medicine are often made by n, reactions.
Transuranium ElementsTransuranium Elements
Elements beyond 92 Elements beyond 92 (transuranium)(transuranium) made made
starting with an starting with an n, n, reaction reaction
2382389292U + U + 11
00n n 2392399292U + U +
2392399292U U 239239
9393Np + Np + 00-1-1
2392399393Np Np 239239
9494Pu + Pu + 00-1-1
47
Q. 56Fe when bombarded with deuterium, produces 54Mn and one other particle. Write a balanced equation for the reaction & identify the other particle.
5626He + 2
1H 5425Mn + ?
48
Nuclide Half-life Organ/SystemIodine-131 8.1 days thyroidIron-59 45.1 days red blood cellsMolybdenum-99 67 hours metabolismPhosphorus-32 14.3 days eyes, liverStrontium-87 2.8 hours bonesTechnetium-99 6 hours heart, bones, liver,
lungs
Medical Uses of Radioisotopes
49
Nonmedical Uses of Radioactive Isotopes
• smoke detectorsAm-241smoke blocks ionized air, breaks circuit
• insect controlsterilize males
• food preservation• radioactive tracers
follow progress of a “tagged” atom in a reaction
50
• authenticating art object many older pigments and ceramics were made from minerals
with small amounts of radioisotopes
• crime scene investigation• measure thickness or condition of industrial
materialscorrosion track flow through processgauges in high temp processes weld defects in pipelines road thickness
Nonmedical Uses of Radioactive Isotopes