Chapter 20 Nuclear Chemistry Nuclear Reactions Differ from chemical reactions One of the most important is the reaction rate is not affected by temperature pressure or the presence of catalysts Take place in the nucleus Occur to lower the energy stabilize of the nucleus Notation 12 6 C 12 mass number protons neutrons 6 atomic number protons C atomic symbol for Carbon Example of an isotope same number of protons different number of neutrons for Carbon 14 Also can be called C 14 1 1 p 0 n 6 C Proton Neutron 1 Electron 0 Positron 0 1 e 0 1 1 e 0 1 2 He Alpha Particle Gamma Rays 4 0 2 4 0 Example Solve for X 212 84 Po 208 82 Pb X Step 1 react mass prod mass 212 208 mass of X Use algebra to find that X mass 4 Step 2 react atomic prod atomic 84 82 atomic of X Use algebra to find that X atomic 2 Step 3 Find what nucleon has the matching terms for X Solution 212 84 Po 208 82 Pb 4 2 or 4 2 He Nuclear Stability Belt of Stability The nuclei that lie in the belt of stability are the most stable isotopes because they have the most optimal ratio of neutrons to protons n p Elements with an atomic greater than 83 does not have any stable isotopes All isotopes want to get to the belt of stability Isotopes below the belt Too little neutrons Too many protons Goal p n To achieve this goal an element can either undergo e capture or positron emission e capture An electron from an orbital combines with a proton from the nucleus to form a neutron 1 1 p 0 1 p 0 1 e or 0 1 e or 0 1 1 1 1 Positron emission A proton can separate into a neutron and a positron 1 0 n 0 n Isotopes above the belt Too many neutrons Too little protons Goal n p To achieve this goal an element undergoes beta particle emission Beta emission A neutron separates into a proton and an electron 1 0 n 0 1 e or 0 1 1 1 p Example 20 11 Na What is the predicted mode of decay Step 1 Find the amount of neutrons and protons 11 protons 20 11 9 neutrons Too little neutrons Step 2 Think about what needs to be done in order to balance out the ratio p n Step 3 Use appropriate method in this case it is e capture or positron emission Solution Positron emission or e capture are appropriate methods Radiation Energy Levels Short wavelength high energy Long wavelength low energy In order from decreasing to increasing energy Types of Radioactive Decay 1 particle emission Occurs when elements with an atomic greater than 83 want to become more stable because emission releases neutrons and protons 239 2 He or 4 2 235 94 Pu 4 92 U 2 particle emission Occurs when elements are above the belt of stability n p Nuclear charge lowers 131 53 I 0 1 e or 0 1 131 54 Xe 3 Positron emission Occurs when elements are below the belt of stability p n Nuclear charge lowers 1 1 p 1 0 n 0 1 e Because of their short half life positrons combine with electrons to emit gamma rays 0 1 e 0 1 e 0 0 4 e capture Occurs when elements are below the belt of stability Nuclear charge lowers p n 1 1 p 0 1 e 1 0 n 5 Gamma rays Emitted when you have other forms of decay 0 0 Nuclear Stability Intermediate elements not too big not too small are the most stable Fe is the most stable element in terms of NBE nucleon Nuclear Binding Energy NBE energy required to break a nucleus apart into its separate pieces p n The higher the NBE the more stable the nucleus E m c2 E NBE m mass in kg c speed of light 3 0 x 108 kg m s Mass defect the mass difference between the sum of the reactants minus the sum of the products prod mass react mass 4 2 He 2 1 1 p 2 1 0 n The mass defect is the difference between the helium and the protons and the neutrons The loss gain in mass is accounted for by energy conversion Example What is the NBE of 4 2 He 4 2 He 2 1 1 p 2 1 0 n Step 1 Find mass defect m prod mass react mass Step 2 Usually mass are given in amu so you must convert amu kg amu x g 6 02 x 1023 amu x kg 103 g Step 3 Plug values into E m c2 This is the energy required to break apart one 4 2 He nucleus Step 4 Find the bonding energy per nucleon E of nucleons In this case it would be 4 nucleons Step 5 Find the bonding energy per mole of He gas Bonding energy per nucleon x 6 02 x 1023 nuclei mole He Kinetics of Radioactive Decay Mechanisms are always unimolecular 1st order process Decay Rate kN k rate constant N of decaying nuclei decay rate is given in disintegrations s Integrated Rate Law ln N t kt ln N 0 t1 2 0 693 k Half Life amount of time it takes for half of the radioactive matter to decay Not affected by temperature or pressure No activation energy EA in nuclear decay Radioactive Dating C 14 dating works on objects 50 000 years or younger Example A wooden artifact is found to have a C 14 activity of 9 1 disintegrations s A piece of new wood has an activity value of 15 2 disintegrations s The half life of C 14 is 5 715 years What is the age of the artifact Step 1 Use equation ln N t kt ln N 0 and t1 2 0 693 k Step 2 Find k using t1 2 0 693 k 5 715 0 693 k Step 3 Plug in values ln 9 1 1 21 x 10 4 yr 1 t ln 15 2 Step 4 Solve for t Solution The artifact is about 4240 years old Nuclear Fusion Nuclear Fission Nuclear fusion Occurs when small nuclei combine to form a larger nucleus This helps stabilize the nucleus Nuclear fission Occurs when a large nucleus separates into smaller nuclei This helps stabilize the nucleus