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Nuclear Chemistry

Nuclear Chemistry. Nuclear Chemistry: Study of nuclear reactions and their applications Nuclear Reactions: changes of the nucleus of atoms Section 21.1 Radioactivity Nucleons : protons , neutrons atomic number : number of protons in the nucleus

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Nuclear Chemistry

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  1. Nuclear Chemistry Nuclear Chemistry: Study of nuclear reactions and their applications Nuclear Reactions: changes of the nucleus of atoms Section 21.1 Radioactivity Nucleons : protons , neutrons atomic number : number of protons in the nucleus mass number = atomic number + number of neutrons isotopes : atoms of the same element which have different mass numbers due to different numbers of neutrons in the nucleus.

  2. Example: where the superscript is the mass number of the isotope. Different isotopes have different natural abundance Different isotopes have different nuclear stability. Nuclide: an atom of a specific isotope. Radioactivity: phenomenon by which an unstable nuclide gets transformed into a more stable one while emitting electromagnetic radiation and other particles.

  3. Types of Radioactive Decay • Alpha Decay ( decay) : The particles emitted during the decay process are -particles : • -particles are He-4 nuclei with a +2 charge. Thus the atomic number of an alpha particle is 2. Its mass number is 4. It is represented either as • When a radioactive nuclide undergoes -decay, the products are -particle and a nuclide which has a mass number 4 less than that of the parent atom and an atomic number 2 less than that of the parent atom. • For example the products of the -decay of 238U are an alpha particle and 234Th :

  4. 2. Beta Decay : The particles emitted during the decay process are -particles . -particles are nothing other than high-speed electrons. They are represented either as When a radioactive nuclide undergoes -decay, the products are -particle and a nuclide which has a mass number equal to that of the parent atom and an atomic number 1 more than that of the parent atom. Beta-decay or beta- emission is equivalent to converting a neutron to a proton and an electron

  5. Gamma radiation or gamma rays are high energy photons. Gamma radiation changes neither the atomic number nor the mass number of a nucleus and is represented as Positron emission : In this type of decay a positron is one of the particles emitted. A positron, is a particle that has the same mass as an electron but an opposite charge ( + 1). The product nuclei will have the same mass number as the parent atom but an atomic number one less than the atomic number of the parent atom.

  6. 5 Electron Capture. This form of decay involves the capture of an inner-shell electron by the nucleus . This process results in the atomic number of the product nuclide being one less than that of the parent while the mass number is the same as that of the parent nuclide.

  7. Patterns of Nuclear Stability Nuclear stability is very complex. In the Plot of no. of protons vs. No. of neutrons , the stable nuclides have in general a neutron to proton ratio within a range called the band of stability. The predictions of nuclear stability based on the belt of stability is based on empirical observations . For nuclei with Z about 20, the ratio 1 for stable nuclei. As Z increases , the ratio increases for stable nuclides. All nuclides with Z  84 are unstable.

  8. Nuclei with 2, 8, 20, 28, 50 or 82 protons or 2,8,20, 28, 50, 82 or 126 neutrons are generally more stable than nuclei that do not contain these numbers of nucleons. These numbers of neutrons and protons are called magic numbers. Nuclei with even numbers of both protons and neutrons are generally more stable than those with odd numbers of nucleons. Patterns of Nuclear Stability Contd.

  9. Predictions Based on Belt-of-stability • Nuclei above belt of stability ( highratio) : • These can decrease the ratio by increasing the number of protons and decreasing the number of neutrons which is achieved by beta emission where a neutron is converted to a proton and an electron. • 2. Nuclei below belt of stability ( lowratio) : • These can increase their ratio by two ways: • positron emission which involves converting a proton into a neutron and a positron • This is common among lighter nuclei; e.g. carbon-11

  10. Predictions Based on Belt of Stability b. Electron capture which involves the capture of an inner shell electron by the nucleus. a.This is more common with heavier nuclei e.g. Rubidium-81 3. Nuclei with Z  84 : These tend to undergo alpha decay. This mode of decay reduces the mass number, A, by 4 and also the atomic number, Z, by 2, moving the unstable nuclei towards the belt of stability. Exceptions: Thorium-233 ( expect alpha observe beta ) Nd-147 ( expect it to be stable but it is radioactive )

  11. Radioactive Series Some nuclei like U-238 cannot gain stability by a single decay since the product nucleus ( alpha decay to Th-234 ) may be unstable as well. So this in turn will undergo further decay ( beta emission to give Pa-234) which also is unstable and hence will also further decay till finally a stable nucleus Pb-206 is formed. Such a series of nuclear reactions that begins with an unstable nucleus and ends in a stable one is known as a radioactive series or a nuclear disintegration series. 238U234Th 234Pa  234U  230Th 226Ra 222Rn  218Po  214Pb 214Bi  214Po  210Pb  210Bi  210Po  206Pb

  12. Nuclear Transmutations Nuclear Reactions in which a nucleus changes identity after being struck by another nucleus or by a neutron. target nucleus : bombarding particle : ejected particle : product nucleus : The same reaction can be showed by a shorthand notation by listing in order the target nucleus, the bombarding particle, the ejected particle and the product nucleus as follows : Target nucleus ( bombarding particle, ejected particle )product nucleus :

  13. a.a. Using Charged Particles The higher the nuclear charge on either the projectile or the target, the higher the projectile speed required to bring about a nuclear reaction. High projectile speeds are achieved by using particle accelerators such as the cyclotron and the synchroton. Strong magnetic and electric fields along the devices combine to accelerate the projectiles to the high speeds required. Artificial heavy elements have been synthesized mainly by this method.

  14. a.b. Using Neutrons When the projectiles are neutrons, there is no need to accelerate them as they are neutral and hence won’t be repelled by the nucleus. The neutrons used are produced by reactions occurring in nuclear reactors. Transuranium Elements Elements with Z > 92 are referred to as the transuranium elements. Artificial transmutations have been used to produce many of these elements.

  15. Rates of Radioactive Decay It is a first order process. The rate of decay is called its activity,A. The S.I unit of activity is the becquerel, Bq 2 Bq = 1 disintegration/s The activity or rate of decay of a radioactive sample is proportional to the number of radioctive nuclei N in the sample : rate =A= kN The constant k is called the decay constant. All radioactive nuclei have a characteristic half-life, t1/2 , that depends on the decay constant, k : For instance, the half-life of C-14 is 5715 yr.

  16. Activity as a Function of Time At = activity after time t Ao = initial activity Nt = number of radioactive nuclei after time t N0 = number of radioactive nuclei initially present

  17. Radioactive Dating C-14 is formed in the upper atmosphere by neutron bombardment by N-14 The ratio of C-14 to C-12 is assumed to have been constant for at least 50,000 years. C-14 is radioactive and undergoes beta decay to N-14. In a living plant or animal the ratio of C-14 to C-12 is constant and identical with the ratio in the atmosphere. Once the organism dies the ratio of C-14 to C-12 decreases with time and the ratio gives an estimate of the age of the organism.

  18. Calculate the decay constant of Cs-90 nuclide which has a half-life of 30 yrs: Answer = 0.693/30 years = 0.023 yr-1 • In-120 undergoes spontaneous radioactive decay by the mode of beta-decay to form which nuclei? 120In  120Sn • Cr-50 undergoes spontaneous radioactive decay by positron emission to form which other nucleus? 50V • Hg-188 undergoes spontaneous decay to form Au-188 and which other particle? Positron • Pu-242 undergoes alpha decay to form what nucleus? 238U • Ta-179 undergoes electron capture to what nucleus ? 179Hf • The particle with the same mass as the electron and with a charge equal in magnitude but opposite in sign to that of the electron is the : positron • alpha particles are identical to : Helium-4 nuclei • Th-234 undergoes spontaneous decay to form Pa-234 and which other particle? Beta particle • 10. The alpha emission by lead-204 results in the product isotope: 200Hg

  19. The half-life of H-3 is 12.3 yr. How much of a 48.0 mg sample of H-3 will remain after 49.2 yr? • Answer: • t1/2 = 12.3 yr; k= 0.693/12.3 yr = 0.0563 yr-1 • ln(mt) = ln(mo) – kt = ln(48.0) - 0.0563 yr-1× 49.2 yr = 1.10 • mt = e1.10 = 3.00 mg • 2. The 14C activity of some ancient Peruvian corn was found to be 10 disintegrations per minute per gram (dpm/g) of carbon. If present-day plant life shows 15 dpm/g, how oldis the Peruvian corn? The half-life of 14C is 5730 yr. • Answer: • t1/2 = 5730 yr; k= 0.693/5730 yr • ln(At) = ln(Ao) – kt ; ln(10) = ln(15) - 0.693/5730 yr × t • t=3350 yr

  20. 3. When a wine was freshly bottled it contained 10 mol of H-3 (tritium). Estimate the current age of the bottled wine if it contains at present 6.0 mol H-3. Tritium decays by beta decay and has a half-life of 12.3 yr. Answer: t1/2 = 12.3 yr; k= 0.693/12.3 yr ln(Nt) = ln(No) – kt ; ln(6) = ln(10) - 0.693/12.3 yr × t 0.693/12.3 yr × t = ln(10)-ln(6) = 0.511 t=(12.3/0.693) ×0.511 yr = 9.07 yr 4. The half-life of Sr-90 1s 28.1 years. How long will it take a 10.0 g sample of Sr-90 to decompose to 0.10 g? Answer t1/2 = 28.1 yr; k= 0.693/28.1 yr ln(mt) = ln(mo) – kt ; ln(0.10) = ln(10.0) – (0.693/28.1 yr) × t (0.693/28.1 yr) × t = ln(10)-ln(0.10) = 4.61 t=(28.1/0.693) ×4.61 yr = 187 yr

  21. A radioactive element has a half-life of 1.0 hr. How many hours will it take for the number of atoms present to decay (decreased or reduced ) by 6.25% of the initial value? Answer: t1/2 = 1.0 hr; k= 0.693/1.0 hr N0 =100; Nt =100-6.25 = 93.75 ln(Nt) = ln(No) – kt ; ln(93.75) = ln(100) - 0.693 hr-1 × t 0.693 ×t hr-1 =ln(100)-ln(93.75) = 0.0645 t=(1/0.693) ×0.0645 hr = 0.0931 hr An ancient wooden artifact was found to have a carbon 14 content 78% of that found in living trees. The half-life of carbon-14 is 5730 years. What is the actual age of the wooden artifact? t1/2 = 5730 yr; k= 0.693/5730 yr N0 =100; Nt =78 ln(Nt) = ln(No) – kt ; ln(78) = ln(100) – (0.693/5730) yr-1 × t (0.693/5730) ×t yr-1 =ln(100)-ln(78) = 0.248 t=(5730/0.693) ×.248 yr = 2050 yr

  22. The half-life for the beta-decay of cesium-137 is 30 years. How many years must pass to reduce a 500 mg sample of cesium-137 to one-millionth of its initial amount? t1/2 =30 yr; k= 0.693/30 yr m0 =500 mg; mt =500 × 10-6 ln(mt) = ln(mo) – kt ; ln(500 × 10-6) = ln(500) – (0.693/30) yr-1 × t (0.693/30) ×t yr-1 =ln(500)-ln(500 × 10-6) = 13.8 t=(30/0.693) ×13.8yr = 598 yr I -131 has a half-life of 8.04 days. Assuming you start with a 5.00 mg sample of I-131, how much will remain after one year? t1/2 =8.04 d; k= 0.693/8.04 d m0 =5.00 mg; mt =? ln(mt) = ln(mo) – kt = ln(5.00) – (0.693/8.04) d-1 × 365 = -29.9 mt =e-29.9 mg = 1.09 ×10-13 mg

  23. 1. Write the balanced nuclear equation for the process summarized as: • 2. Using the shorthand notation, write the nuclear reaction • 3. Which nuclear equation is properly balanced? • A. 4He + 24Mg 27Si + 1H (atomic numbers do not add to give the same number) • B. 14 N + 0 –1e 14O (atomic numbers do not add to give the same number) • 4He + 9Be 12 C + 1H • D. 4He + 14N 17O + 1H

  24. 1. In the following nuclear equation, identify the missing particle. 4320 Ca + 42 He  ____ + 11 H Answer: 4621Sc

  25. The mass changes, m, and the associated energy changes, E, in nuclear reactions are much greater than those in chemical reactions. m = total mass of the products – total mass of the reactants. The energy change per mole associated with a nuclear reaction is calculated using Einstein’s equation : If there is a loss of mass (m<0) accompanying the reaction, the reaction will be exothermic (E<0) . If there is a gain of mass (m>0) accompanying the reaction, the reaction will be endothermic (E>0) . All spontaneous nuclear reactions are exothermic. Energy Changes in Nuclear Reactions

  26. 1. One of the hopes for solving the world’s energy problem is to make use of the fusion reaction: 21 H + 31 H 42 He + 10 n + energy (2.01400g) (3.01605g) (4.002603g) (1.008665g)  how much energy is released when one mole of deuterium is fused with one mole of tritium according to the above reaction? The speed of light is 2.9979x108 m/s Answer: m for 1 mol of = -0.018797 g = -0.000018797 kg E = m×c2 = -0.000018797 kg×(2.9979x108 m/s)2 = -1.69 × 1012 J 2. Calculate the energy change per mole accompanying the following nuclear reaction Nuclear mass of U-235 : 234. 9359 amu; Nuclear mass of Ba-141 : 140.8833 amu; Nuclear mass of Kr-92 : 91.9021 amu Mass of a neutron : 1.0087 amu m for 1 mol of = -0.1331 g = -0.0001331 kg E = m×c2 = -0.0001331 kg×(2.9979x108 m/s)2 = -1.20 × 1013 J

  27. Nuclear Binding Energies Mass Defect of a nucleus, m : difference between the mass of a nucleus and its constituent nucleons. Nuclear Binding Energy, E : energy required to separate a nucleus into its individual nucleons. The larger the binding energy per nucleon, more stable the nucleus. Nuclei of intermediate mass numbers are more tightly bound than those with smaller or larger mass numbers. This is why heavy nuclei gain stability (lose energy) by breaking into two mid-sized nuclei (nuclear fission). This also explains why nuclear fusion of very light nuclei releases energy.

  28. Calculate the binding energy per nucleon of an Al-27 nucleus given the following masses: Al-27 nucleus : 26.981541 amu; proton : 1.007838 amu; neutron : 1.008665 amu Answer: E = m×c2 = m= sum of masses of nucleons – mass of the nucleus Al-27 has 13 p, 14 n sum of masses of nucleons = 13× 1.007838 + 14× 1.008665 = 27.223204 amu m= 27.223204 amu - 26.981541 amu = 0.241663 amu = 0.241663 amu × (1g/6.022x1023 amu) = 4.013x10-25 g = 4.013x10-28 kg Binding energy, E= 4.013x10-28 kg×(2.9979x108 m/s)2 =3.6x10-11 J Binding energy per nucleons = 3.6x10-11 J/27 nucleons = 1.3x10-12 J/nucleon

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