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Chapter 21 Nuclear Chemistry
© 2012 Pearson Education, Inc.
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© 2012 Pearson Education, Inc.
The Nucleus Remember that the nucleus is composed of the two nucleons, protons and neutrons. The number of protons is the atomic number. The number of protons and neutrons together is effectively the atomic mass. © 2012 Pearson Education, Inc.
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© 2012 Pearson Education, Inc.
Isotopes Not all atoms of the same element have the same mass, due to different numbers of neutrons in those atoms. There are, for example, three naturally occurring isotopes of uranium: Uranium-234 Uranium-235 Uranium-238 © 2012 Pearson Education, Inc.
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© 2012 Pearson Education, Inc.
Radioactivity It is not uncommon for some nuclides of an element to be unstable, or radioactive. We refer to these as radionuclides. There are several ways radionuclides can decay into a different nuclide. © 2012 Pearson Education, Inc.
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Types of Radioactive Decay
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© 2012 Pearson Education, Inc.
Alpha Decay Alpha decay is the loss of an α-particle (a helium nucleus): He 4 2 U 238 92 Th 234 90 He 4 2 + © 2012 Pearson Education, Inc.
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Sample Exercise 21.1 Predicting the Product of a Nuclear Reaction
What product is formed when radium-226 undergoes alpha decay? Practice Exercise Which element undergoes alpha decay to form lead-208?
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+2 –2 +4 –4 Answer: D
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Beta Decay Beta decay is the loss of a -particle (a high-energy electron): 1 e or I 131 53 Xe 54 + e 1 © 2012 Pearson Education, Inc.
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Electron Capture (K-Capture)
Addition of an electron to a proton in the nucleus is known as electron capture or K-capture. The result of this process is that a proton is transformed into a neutron: p 1 + e 1 n © 2012 Pearson Education, Inc.
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© 2012 Pearson Education, Inc.
Positron Emission Some nuclei decay by emitting a positron, a particle that has the same mass as, but an opposite charge to, that of an electron: e 1 C 11 6 B 5 + e 1 © 2012 Pearson Education, Inc.
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© 2012 Pearson Education, Inc.
Gamma Emission Gamma emission is the loss of a -ray, which is high-energy radiation that almost always accompanies the loss of a nuclear particle: © 2012 Pearson Education, Inc.
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Sample Exercise 21.2 Writing Nuclear Equations
Write nuclear equations for (a) mercury-201 undergoing electron capture; (b) thorium-231 decaying to protactinium-231. Practice Exercise Write a balanced nuclear equation for the reaction in which oxygen-15 undergoes positron emission. 13
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Alpha particle Neutron Proton Beta particle Answer: B
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Neutron–Proton Ratios
Any element with more than one proton (i.e., anything but hydrogen) will have repulsions between the protons in the nucleus. A strong nuclear force helps keep the nucleus from flying apart. Neutrons play a key role stabilizing the nucleus. Therefore, the ratio of neutrons to protons is an important factor. For smaller nuclei (Z 20), stable nuclei have a neutron-to-proton ratio close to 1:1. As nuclei get larger, it takes a larger number of neutrons to stabilize the nucleus. © 2012 Pearson Education, Inc.
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© 2012 Pearson Education, Inc.
Stable Nuclei The shaded region in the figure, the so-called belt of stability, shows what nuclides would be stable. Nuclei above this belt have too many neutrons. These nuclei tend to decay by emitting beta particles. Nuclei below the belt have too many protons. Nuclei tend to become more stable by positron emission or electron capture. © 2012 Pearson Education, Inc.
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© 2012 Pearson Education, Inc.
Stable Nuclei There are no stable nuclei with an atomic number greater than 83. Nuclei with such large atomic numbers tend to decay by alpha emission. © 2012 Pearson Education, Inc.
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Sample Exercise 21.3 Predicting Modes of Nuclear Decay
Predict the mode of decay of (a) carbon-14, (b) xenon-118. Practice Exercise Predict the mode of decay of (a) plutonium-239, (b) indium-120. 18
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A. 82 neutrons B. 92 neutrons C. 102 neutrons D. 112 neutrons
Answer: C
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Radioactive Series Large radioactive nuclei cannot stabilize by undergoing only one nuclear transformation. They undergo a series of decays until they form a stable nuclide (often a nuclide of lead). © 2012 Pearson Education, Inc.
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© 2012 Pearson Education, Inc.
Some Trends Nuclei with 2, 8, 20, 28, 50, or 82 protons or 2, 8, 20, 28, 50, 82, or 126 neutrons tend to be more stable than nuclides with a different number of nucleons. Nuclei with an even number of protons and neutrons tend to be more stable than nuclides that have odd numbers of these nucleons. © 2012 Pearson Education, Inc.
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A. Even number of protons and fewer than three stable isotopes, 2 elements; odd number of protons and more than two isotopes, 2 elements B. Even number of protons and fewer than three stable isotopes, 1 element; odd number of protons and more than two isotopes, 3 elements Even number of protons and fewer than three stable isotopes, 4 elements; odd number of protons and more than two isotopes, 1 element Even number of protons and fewer than three stable isotopes, 3 elements; odd number of protons and more than two isotopes, 0 elements Answer: D
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All have an even number of neutrons.
Answer: A All have an even number of neutrons. All have an odd number of neutrons. F and Na have an odd number of neutrons whereas Al and P have an even number F and P have an odd number of neutrons whereas Al and Na have an even number.
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Practice Exercise Sample Exercise 21.4 Predicting Nuclear Stability
Predict which of these nuclei are especially stable: , , Practice Exercise Which of the following nuclei would you expect to exhibit a special stability: , , ? . 24
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Nuclear Transformations
Nuclear transformations can be induced by accelerating a particle and colliding it with the nuclide. © 2012 Pearson Education, Inc.
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Particle Accelerators
These particle accelerators are enormous, having circular tracks with radii that are miles long. © 2012 Pearson Education, Inc.
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Sample Exercise 21.5 Writing a Balanced Nuclear Equation
Write the balanced nuclear equation for the process summarized as Practice Exercise Write the condensed version of the nuclear reaction 27
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Yes, because a neutron has mass and can interact with an electrostatic or magnetic field and is accelerated by an electrostatic or magnetic field. No, because a neutron has no charge and therefore cannot be accelerated by an electrostatic or magnetic field. Answer: B
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Kinetics of Radioactive Decay
Nuclear transmutation is a first-order process. The kinetics of such a process, you will recall, obey this equation: = −kt Nt N0 ln © 2012 Pearson Education, Inc.
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Kinetics of Radioactive Decay
The half-life of such a process is = t1/2 0.693 k Comparing the amount of a radioactive nuclide present at a given point in time with the amount normally present, one can find the age of an object. © 2012 Pearson Education, Inc.
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A g B g C g D g Answer: C
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Sample Exercise 21.6 Calculation Involving Half-Lives
The half-life of cobalt-60 is 5.3 yr. How much of a mg sample of cobalt-60 is left after 15.9 yr? Practice Exercise Carbon-11, used in medical imaging, has a half-life of 20.4 min. The carbon-11 nuclides are formed, and the carbon atoms are then incorporated into an appropriate compound. The resulting sample is injected into a patient, and the medical image is obtained. If the entire process takes five half-lives, what percentage of the original carbon-11 remains at this time? 32
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Measuring Radioactivity
One can use a device like this Geiger counter to measure the amount of activity present in a radioactive sample. The ionizing radiation creates ions, which conduct a current that is detected by the instrument. © 2012 Pearson Education, Inc.
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Kinetics of Radioactive Decay
A wooden object from an archeological site is subjected to radiocarbon dating. The activity of the sample that is due to 14C is measured to be 11.6 disintegrations per second. The activity of a carbon sample of equal mass from fresh wood is 15.2 disintegrations per second. The half-life of 14C is 5715 yr. What is the age of the archeological sample? © 2012 Pearson Education, Inc.
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Kinetics of Radioactive Decay
First we need to determine the rate constant, k, for the process: = t1/2 0.693 k = 5715 yr 0.693 k = k 0.693 5715 yr = k 1.21 104 yr1 © 2012 Pearson Education, Inc.
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Kinetics of Radioactive Decay
Now we can determine t: = −kt Nt N0 ln = −(1.21 104 yr1)t 11.6 15.2 ln = −(1.21 104 yr1)t ln 0.763 = t 2240 yr © 2012 Pearson Education, Inc.
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Spontaneous radioactive processes have differing reaction mechanisms that do not fit a zero-order or second-order kinetic process. Spontaneous radioactive processes have differing reaction rates that do not fit a zero-order or second-order kinetic process. Spontaneous radioactive processes are either unimolecular or trimolecular processes leading to a first-order or third-order kinetic process. Spontaneous radioactive processes are unimolecular processes leading only to a first-order kinetic process. Answer: D
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(a) Yes (see equation 21.18) (b) No (see equation 21.20)
(a) No (see equation 21.18) (b) Yes (see equation 21.20) (a) No (see equation 21.18) (b) No (see equation 21.20) (a) Yes (see equation 21.18) (b) Yes (see equation 21.20) Answer: A
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Sample Exercise 21.7 Calculating the Age of a Mineral
A rock contains mg of lead-206 for every milligram of uranium-238. The half-life for the decay of uranium-238 to lead-206 is 4.5 × 109 yr. How old is the rock? Practice Exercise A wooden object from an archeological site is subjected to radiocarbon dating. The activity due to 14C is measured to be 11.6 disintegrations per second. The activity of a carbon sample of equal mass from fresh wood is 15.2 disintegrations per second. The half-life of 14C is 5715 yr. What is the age of the archeological sample? 39
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Sample Exercise 21.8 Calculations Involving Radioactive Decay
If we start with g of strontium-90, g will remain after 2.00 yr. (a) What is the half-life of strontium-90? (b) How much strontium-90 will remain after 5.00 yr? (c) What is the initial activity of the sample in becquerels and curies? Practice Exercise A sample to be used for medical imaging is labeled with 18F, which has a half-life of 110 min. What percentage of the original activity in the sample remains after 300 min? 40
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Yes No. Alpha and beta particles will pass through matter with equal efficiency but not gamma rays No. Alpha particles and gamma rays will pass through matter with equal efficiency but not beta particles. No. Alpha particles are more readily absorbed by matter than beta particles and gamma rays. Answer: D
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Energy in Nuclear Reactions
There is a tremendous amount of energy stored in nuclei. Einstein’s famous equation, E = mc2, relates directly to the calculation of this energy. In the types of chemical reactions we have encountered previously, the amount of mass converted to energy has been minimal. However, these energies are many thousands of times greater in nuclear reactions. © 2012 Pearson Education, Inc.
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Energy in Nuclear Reactions
For example, the mass change for the decay of 1 mol of uranium-238 is g. The change in energy, E, is then E = (m)c2 E = (4.6 106 kg)(3.00 108 m/s)2 E = 4.1 1011 J © 2012 Pearson Education, Inc.
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Sample Exercise 21.9 Calculating Mass Change in a Nuclear Reaction
How much energy is lost or gained when 1 mol of cobalt-60 undergoes beta decay, ? The mass of a atom is amu, and that of a atom is amu. Practice Exercise Positron emission from 11C, , occurs with release of 2.87 × 1011 J per mole of 11C. What is the mass change per mole of 11C in this nuclear reaction? The masses of 11B and 11C are and amu, respectively. 44
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The atomic mass of iron-56 is the weighted average of the naturally occurring isotopes of iron-56, not just the mass of the nucleus. The values in Table 21.7 reflect only the mass of the nucleus while the atomic mass of iron-56 also includes the mass of its neutrons. The values in Table 21.7 reflect only the mass of the nucleus while the atomic mass of iron-56 also includes the mass of its electrons. The values in Table 21.7 reflect only the mass of the nucleus while the atomic mass iron-56 also includes the mass of its protons. Answer: C
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Yes, nuclei having mass numbers around 100 are not very stable nuclei.
No, nuclei have mass numbers around 100 are among the most stable of nuclei. Answer: B
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Nuclear Fission How does one tap all that energy? Nuclear fission is the type of reaction carried out in nuclear reactors. © 2012 Pearson Education, Inc.
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© 2012 Pearson Education, Inc.
Nuclear Fission Bombardment of the radioactive nuclide with a neutron starts the process. Neutrons released in the transmutation strike other nuclei, causing their decay and the production of more neutrons. This process continues in what we call a nuclear chain reaction. © 2012 Pearson Education, Inc.
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© 2012 Pearson Education, Inc.
Nuclear Fission Therefore, there must be a certain minimum amount of fissionable material present for the chain reaction to be sustained: critical mass. © 2012 Pearson Education, Inc.
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© 2012 Pearson Education, Inc.
Nuclear Reactors In nuclear reactors, the heat generated by the reaction is used to produce steam that turns a turbine connected to a generator. © 2012 Pearson Education, Inc.
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B. To provide a water pond for storage of nuclear waste
Answer: A A. To provide sufficient water to condense the secondary coolant after it passes through a turbine B. To provide a water pond for storage of nuclear waste C. To provide steam to heat the reactor D. To provide sufficient coolant to cool the entire operating environment containing the nuclear reactor
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Country: Most Reactors Most Reactors Highest % Operating Construction
Answer: B Country: Most Reactors Most Reactors Highest % Operating Construction A. China United States Germany B. United States China France C. France China Russia D. China France Canada
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Nuclear Reactors The reaction is kept in check by the use of control rods. These rods block the paths of some neutrons, keeping the system from reaching a dangerous supercritical mass. © 2012 Pearson Education, Inc.
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© 2012 Pearson Education, Inc.
Nuclear Fusion Fusion would be a superior method of generating power. The good news is that the products of the reaction are not radioactive. The bad news is that in order to achieve fusion, the material must be in the plasma state at several million kelvins. © 2012 Pearson Education, Inc.
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A. Alpha rays are dangerous only when the alpha emitter is near cells inside the body.
B. Alpha rays do not penetrate the exterior skin effectively but once inside the body they are dangerous to cells. C. Alpha rays need to be close to blood cells to be damaging but other cells inside the body are far less affected. D. Alpha rays can penetrate the walls of cells inside the body only when the alpha emitter is in direct contact with the cells. Answer: B
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Absorbed dose, 3 rads; effective dosage, 33 rems.
Answer: C
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