Nuclear Chemistry Chapter 23 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. PowerPoint Lecture Presentation by J. David Robertson University of Missouri

REACTIONS CHEMICALNUCLEAR RADIOACTIVE DECAY Unstable nuclei emits particles and/or electromagnetic radiation spontaneously. NUCLEAR TRANSMUTATION Bombardment of nuclei by neutrons, protons or other nuclei; occurrence are either naturally or artificially.

23.1

X A Z Mass Number Atomic Number Element Symbol 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 1p1p 1 1H1H 1 or proton 1n1n 0 neutron 0e0e 00 or electron 0e0e +1 00 or positron 4 He 2 44 2 or  particle

Radioactive Decay 23.2 A stable nucleus remains intact, but the majority of nuclei are unstable. An unstable nucleus exhibits radioactivity – it spontaneously disintegrates or decays, by emitting radiation. Each type of unstable nucleus has a characteristic rate of radioactive decay – might range from a fraction of a second to several billion years. “when a nuclide of one element decays, it changes into a nuclide of a different element”

When a nuclide decays, it forms a nuclide of lower energy, and the excess energy is carried off by the emitted radiation. The decaying nuclide is called ‘the parent’ and the product nuclide is called ‘the daughter’.

Type of Radioactive Decay 23.2 Alpha decay: loss of an α particle Beta decay: ejection of a β particle A decreases by 4 Z decreases by 2 A does not change Z increases by 1 Positron decay: emission of a positron A does not change Z decreases by 1 Positron is the ‘antiparticle’ of the electron. Positron decay has the opposite effect of β decay.

Electron capture: Occurs when the nucleus of an atom draws in an electron from the lowest energy level. The symbol is used to distinguished it from a β particle. A does not change Z decreases by 1 Gamma Emission Loss of a  -ray (high-energy radiation) that almost always accompanies the loss of a nuclear particle.

Balancing Nuclear Equations 1.Conserve mass number (A). The sum of protons plus neutrons in the products must equal the sum of protons plus neutrons in the reactants. 1n1n 0 U Cs Rb n1n = (2x1) 2.Conserve atomic number (Z) or nuclear charge. The sum of nuclear charges in the products must equal the sum of nuclear charges in the reactants. 1n1n 0 U Cs Rb n1n = (2x0) 23.1

QUESTION #1: a)What radioactive isotope is produced in the following bombardment of boron? Answer: Note: Identify the element using the atomic number (proton number). Refer to the Periodic Table.

b)Write the nuclear equation for the beta emitter Co–60. Answer: Note: ‘Beta emitter’ means the reaction emits/produces beta particles.

c) Complete the following nuclear reactions: a) 40 Ar( , p)? b) 59 Co(n,  )? Answer: Mn

Question #2: 212 Po decays by alpha emission. Write the balanced nuclear equation for the decay of 212 Po Alpha particle:or Checking the Periodic Table for the element with atomic number (proton number) of 82 gives the answer as Pb. Therefore:

Question #3: In the following reactions, identify X: (a) (b) Answer: (a) (b)

Neutron-proton ratio (n/p ratio): 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. *Nuclear force = the force between nucleons i.e.: protons & neutrons Neutrons play a key role stabilizing the nucleus. Therefore, the ratio of neutrons to protons (n/p) is an important factor in determining the stability of nuclides. Nuclear Stability & Mode of Decay

For smaller nuclei (Z  20) stable nuclei have a neutron-to-proton ratio close to 1:1. As nuclei get larger, it takes a greater number of neutrons to stabilize the nucleus. The shaded region in the figure shows what nuclides would be stable, the so-called belt of stability.

23.2 Nuclei above this belt have too many neutrons. They tend to decay by emitting beta particles. Nuclei below the belt have too many protons. They tend to become more stable by positron emission or electron capture.

Nuclear Stability 23.2 Certain numbers of neutrons and protons are extra stable: Nuclei with n or p = 2, 8, 20, 50, 82 and 126 (magic numbers); Extra stable numbers of electrons in noble gases (e = 2, 10, 18, 36, 54 and 86). Nuclei with even numbers of both protons and neutrons are more stable than those with odd numbers of neutron and protons.

All isotopes of Tc (Technetium) and Pm (Promethium) are radioactive All isotopes of the elements with atomic numbers higher than 83 are radioactive.

Question #4: Which of the following nuclides would you predict to be stable and which is radioactive? Explain. Answer: Radioactive. The n/p = 0.8. Despite having even number of proton and neutron, the minimum ratio for stability is 1.0. This nuclide has too few neutron to be stable. Stable. The n/p = 1.0 and it has even number of proton and neutron. Radioactive. Every nuclide with Z > 83 is radioactive. Radioactive. The n/p = 1.20, therefore this nuclide is not stable.

Nuclear binding energy is the energy required to disassemble a nucleus into free unbound neutrons and protons. Nuclear binding energy can be calculated from the difference of mass of a nucleus, and the sum of the masses of the number of free neutrons and protons that make up the nucleus. This mass difference is called the mass defect or mass deficiency, next, Einstein's formula: E = mc 2 can be used to compute the binding energy of any nucleus. Nuclear Binding Energy

Nuclear binding energy per nucleon vs Mass number nuclear binding energy nucleon nuclear stability 23.2

Calculation Steps of Binding Energy per Nucleon: Step 1: Calculate mass defect. Mass defect is the difference between the mass of an atom and the sum of the masses of its protons and neutrons. Unit is in amu. Step 2: Convert amu to kg. Step 3: Calculate nuclear binding energy, BE. Use the formula: E = mc 2. Step 4: Divide binding energy by number of nucleons to get nuclear binding energy per nucleon.

23.2 Answer: Question #5: Iron-56 is an extremely stable nuclide. Calculate the binding energy (BE) per nucleon for iron-56. Mass of 56 Fe atom = a.m.u Mass of proton = a.m.u Mass of neutron = a.m.u Mass defect,  m = mass product – mass reactant = (mass p + mass n) – mass 56 Fe  m = amu – amu = amu mass p+n = (25 x amu) + (31 x amu) Step 1: Calculate mass defect.

 m = amu x  m = x kg Binding energy, BE:  E = mc 2 = x kg x (3.00 x 10 8 m/s) 2  E = 7.91 x J Binding energy per nucleon: Step 2: Convert amu to kg. Step 3: Calculate nuclear binding energy, BE. Step 4: Divide BE by number of nucleons.

23.2 Answer: Question #6: What is the nuclear binding energy per nucleon for 25 Mg? Mass of 25 Mg = a.m.u Mass of proton = a.m.u Mass of neutron = a.m.u Mass defect,  m = mass product – mass reactant = (mass p + mass n) – mass 25 Mg  m = amu – amu = amu mass p+n = (12 x amu) + (13 x amu)

 m = amu x  m = x kg Binding energy, BE:  E = mc 2 = x kg x (3.00 x 10 8 m/s) 2  E = 3.30 x J Binding energy per nucleon:

For each duration (half-life), one half of the substance decomposes. For example: Ra-234 has a half-life of 3.6 days. If you start with 50 grams of Ra-234: After 3.6 days  25 grams After 7.2 days  12.5 grams After 10.8 days  6.25 grams Half-Life The time taken for ½ of a sample to decompose. The rate of a nuclear transformation depends only on the “reactant” concentration. All radioactive decays obey first order kinetics. Kinetics of Radioactive Decay

N daughter rate = - NN tt rate = N NN tt = N - N = N 0 exp(- t)ln N = ln N 0 - t N = the number of atoms at time t N 0 = the number of atoms at time t = 0 is the decay constant ln2 = t½t½ 23.3

Kinetics of Radioactive Decay [N] = [N] 0 exp(- t) ln[N] = ln[N] 0 - t [N] ln [N] 23.3

Question #7: The half-life of radium-226 is 1.60  10 3 years. How many hours will it take for a 2.50 g sample to decay until g of the isotope remains? Answer: ln N t = ln N 0 - t ln = ln 2.50 – (4.33 x yrs -1 )t  t = 6013 years or 5.27  10 7 hours

Question #8: If 12% of a certain radioisotope decays in 5.2 years, what is the half life of this isotope? Answer: ln N t = ln N 0 - t ln 0.88 = ln 1.00 – (5.2) If N 0 = 1.00 then N t = 1.00 – 0.12 = 0.88  = years –1

The figure shows the 238 U Decay Series. Natural Radioactivity Radioactive decay series is a sequence of nuclear reactions that ultimately results in the formation of a stable isotopes. The beginning radioactive isotope is called the parent and the product is called the daughter.

Natural Radioactivity : Radiocarbon Dating for Determining the Age of Artifacts

Radioactive C-14 is formed in the upper atmosphere by nuclear reactions initiated by neutrons in cosmic radiation 14 N + 1 o n ---> 14 C + 1 H The C-14 is oxidized to CO 2, which circulates through the biosphere. When a plant dies, the C-14 is not replenished. But the C-14 continues to decay with t 1/2 = 5730 years. Activity of a sample can be used to date the sample. Natural Radioactivity : Radiocarbon Dating for Determining the Age of Artifacts

Natural Radioactivity : Dating based on Radioactive Decay Dating using uranium-238 isotopes: Some of the intermediate products in the uranium decay series have very long half-lives, therefore they are suitable for estimating the age of rocks in the earth and of extraterrestrial objects. Dating using potassium-40 isotopes: The radioactive potassium-40 isotope decays by several modes but the important one is of that electron capture. This particular mode can be use to gauge the age of a specimen in geochemistry.