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Chemistry Session opener Nuclear chemistry, sun & life Session objectives 1. Radioactivity 2. Nuclear reactions 3. Kinetics of radioactive decay 4. Radioactivity equilibrium 5. Fission reaction 6. Fusion reaction 7. Radiocarbon dating 8. Medical applications Magic numbers RADIOACTIVITY AND NUCLIDE STABILITY protons neutrons stable nuclides even even 157 even odd 52 odd even 50 odd odd 5 “Magic numbers” analgous to the noble gas electronic configurations occur at: 2, 8, 20, 28, 40, 50, 82, 126, 184 The most stable isotopes have “magic numbers” of both protons and neutrons. n/p too large beta decay X Y n/p too small positron decay or electron capture 23.2 Stability of nucleus It has been observed that for light elements of atomic number upto 20, n/p ratio is 1 When the n/p ratio is too high This nucleus would be unstable and would tend to lower the ratio by emitting -particles and thus moves towards the zone of stability. 1 0n 1 H1 1 e 0 neutrons proton β - particle (electron) Stability of nucleus When the n/p ratio is too low This is effected either by the emission of an alpha particle or a positron or by capturing an orbital electron 11 Na20 10 Ne20 + +1e0 positron 37 Rb82 + -1e0 36 Kr 82 92 U235 90 Th231 + 2He4 particle Nuclear reactions Loss of –particle z Am z2 Bm4 2 He4 ( particle) Loss of –particle z Am z1 Bm 1 e0 Group displacement laws The emission of an -particle results in the formation of an element which lies two places to the left and the emission of a -particle results in the formation of an element which lies one place to the right in the periodic table Illustrative Example 92 U238 is radioactive and emits a and b particles to form 82 Pb206 . What is the respective values of a and b? Solution Let x number of particles and y number of particles get emitted. 92U238 x 2 He 4 y 1e 0 82Pb 206 238 = 4x + 206 or x = 8 92 = 2x – y + 82 or y = 6 Solution Cont. 212Podecays by alpha emission. Write the balanced nuclear equation for the decay of 212Po. 212Po 4He + AX 84 2 Z 212 = 4 + A A = 208 84 = 2 + Z Z = 82 212Po 4He + 208Pb 84 2 82 23.1 Disintegration series Name of the Series Initial element Stable element Value of Value of n n for for the initial stable element element Series 4n Thorium series Thorium–232 Lead–208 58 52 4n + 1 Neptunium series Neptunium– Bismuth–209 59 52 237 4n + 2 Uranium series Uranium–238 Lead–206 59 51 4n + 3 Actinium series Uranium–235 Lead–207 58 51 4n series In the 4n series,all nuclides have mass number that are multiples of 4. 4n +1 series In the 4n + 1 series,all nuclides have mass numbers that are one number greater than multiples of 4. 4n+2 series In the 4n + 2 series, all nuclides have mass number that are two numbers greater than multiples of 4. 4n+3 series In the 4n + 3 series, all nuclides have mass numbers that are three numbers greater than multiples of 4. Nuclear Binding Energy Nuclear binding energy The energy required to break up a nucleus into its component protons and neutrons. BE+19 F 91p+101 n 9 1 0 E = mc2 BE = 9 x (p mass) + 10 x (n mass) – 19F mass BE (amu) = 9 x 1.007825 + 10 x 1.008665 – 18.9984 BE = 0.1587 amu 1 amu = 1.49 x 10-10 J BE = 2.37 x 10-11J Binding energy Binding energy per nucleon= Number of nucleons 2.37 x 10-11 J = =1.25 x 10-12 J 19 nucleons Average Binding Energy as a Function of Atomic Number Kinetics of radioactive decay M D at t = 0 N0 at t = t N N0=Initial no. of radioactive particles N=No. of radioactive particles after time t Kinetics of radioactive decay The rate of disintegration of dN M into D is equal to dt dN dN i.e., N or N (I) dt dt N dN or N dt No No or n t (II) N N Noet (iv) 2.303 N log10 0 (iii) t N Half life period .693 t1 / 2 Average or Mean life Period 1 t1 / 2 1.44 t1 / 2 0.693 Units 1 curie (C) = 3.7 × 1010 dps S.I unit is Beequerel (Bq) (Bq) = 1 disintegration per second. Illustrative example Calculate the activity in terms of dpm of 0.001 g sample of Pu239 (half life = 24300 years). Solution: 0.693 Calculation of decay constant t1 / 2 24300 3.15 107 t1 / 2 minutes 60 0.693 60 min1 24300 3.15 107 103 6.02 1023 Number N of the nuclei in the given sample 239 dN Hence, N dt dN 0.693 60 103 6.023 1023 = 1.37 × 108 dpm. dt 24300 3.15 107 239 Illustrative example One of the hazards of nuclear explosion is the generation of Sr90 and its subsequent incorporation in bones. This nuclide has a half-life of 28.1 years. Suppose 1m g was absorbed by a new born child, how much Sr90 will remain in his bones after 20 years? Solution 0.693 0.693 We have 0.0246 yr 1. t1/ 2 28.1 2.303 a 2.303 a As t log 20 log ax 0.0246 ax a 1.64 ax Here a = 10–6 g a – x = 6.09 × 10–7 g Illustrative example Calculate the weight of Na24 which will give radioactivity of one curie (half-life of Na24 is 15 hr). 1C = 3.7 × 1010 dps Solution 0.693 0.693 Now 1.28 105 t1/ 2 15 60 60 3.7 1010 N .....(i) N = 2.89 × 1015 6.023 × 1023 of Na24 atoms = 24 g 24 2.89 1015 2.89 10 15 of Na atoms 24 1.15 10 7 g 6.023 1023 Illustrative example The half-life for the decay of U238 to Th234 is 4.6 × 109 years. How many a particles are produced per second in a sample containing 3 × 1020 atoms of U238 . Solution 0.693 We know that t1/ 2 0.693 or 1.5 1010 yr 1 4.6 109 dN Now N 1.5 1010 3 1020 dt dN 4.52 1010 atoms yr 1 dt Number of particles emit per second 4.52 1010 1433 365 24 60 60 Artificial transmutation of elements Process of transformation of one element into the other by artificial means i.e. by bombarding the nuclei of atom by high speed subatomic particles is called artificial transmutation of elements: Rutherford 1919. 14N 7 + 4He → 2 17O 8 + 1H 1 Irene Joliot-Curie. 24Al 13 + 4He → 2 30P 15 + 1n 0 15 14 +1 30P → 30Si + 0 Shared Nobel Prize 1938 Transuranium Elements Element coming after uranium [Z = 92] in the periodic table i.e. with atomic number greater than 92 are called transuranic elements. e.g. 238 92 U + 1 0n → 239 + 92 U 239 → 239 93 Np + 0 92 U -1 249 98 Cf + 15 N → 7 260 105 U 1 + 4 0n Fission of U-235 Fission of U-235 56 Ba 36 Kr 31 n 140 93 0 92 U 0 n 92 U 235 1 236 54 Xe 38 Sr 21 n 144 90 0 55 Co 37 Rb 21 n 144 90 0 The minimum amount of fissionable material required to continue a nuclear chain reaction is called critical mass. Nuclear Reactors Breeder Reactors A breeder reactor is one that produces more fissionable nuclei than it consumes i.e. it increases the concentration of fissionable nuclie. 238U 92 + 1 0n → 1 239U 92 → -1 239U 239Np + 0 92 93 → -1 239Np 239Pu 0 93 94 + Fusion 2 2 1 4 H 1 H 2He energy. Difference between fission and fusion Nuclear fission Nuclear fusion 1. It involves breaking 1. It involves fusion of up of a heavier two or more lighter nucleus into lighter nuclei to form a heavier nucles. nucleus. 2. Large number of 2. It is difficult to control radioisotopes are this process. formed. 3. It is a chain 3. It is not a chain process. process. 4. This does not 4. This is initiated by require high very high temperature. temperature 5. It can be controlled 5. It cannot be and energy is controlled. released during the reactions. Rate law for reactions involving parallel reactions K1 B (80%) Main reaction A A (20%) Side reaction K2 d{A} Rate K1[A] K 2 [A] (K1 K 2 )[A] dt Illustrative example 227Ac has a half-life of 22 years with respect to radioactive decay. The decay follows two parallel paths, one leading to 227Th and the other to 223Fr. The percentage yields of these two daughter nuclides are 2 and 98 respectively. What are the decay constants (l) for each of the separate path? Solution 1 2 27 h% ( T 2) 2 c 7 2A 2 2 23 r 8 ( % F9 ) 0.693 On solving, total 0.0315 per year 22 1 6.3 10 4 per year 1 2 0.0315 2 3.08 10 2 per year 2 98 while 48 1 2 Radioactive equilibrium A B C D If the rate of decay of B is the same as its rate of formation from A, then the amount of B will remain constant. dNA dNB dt dt NA B NB A Age of minerals and rocks The end product in the natural disintegration series is an isotope of lead. Each disintegration step has a definite decay control. By finding out the amounts of percent radioactive element and the isotope of lead (e.g. 92U238 and 82Pb206) in a sample of rock and knowing the decay constant of the series, the age of rock can be calculated. Illustrative Example An uranium containing ore pitch blende contains 0.055 g of 82Pb206 for 1 g of 92U238. Calculate the age of the ore. (Half-life = 4.6 × 109 years) Solution 2.303 N t log o , N 0.693 We have 0.15 109 4.6 109 Here N = 1 g 238 Now 0.055 g of Pb206 0.055 gm of U238 0.0635 g 206 No 1 0.0635 1.0635 g 2.303 1.0635 t log 4.10 108 yrs. 0.15 109 1 Radio carbon dating The principle is based upon the formation of C14 by neutron capture in the upper atmosphere. 7 N14 + 0n1 6C14 + 1H1 This carbon-14 is radioactive with a half life period of 5700 years. This is changed into nitrogen by the emission of particles. 14 N14 + e0 t = 5700 yrs 6C 7 -1 1 2 Radio carbon dating The age of the object can be estimated. 2.303 initial activity (amount of C14 in a living object) tiem(t) log Final activity (amount of C14 in a dead object) This method cannot be applied to estimate the age of an object which is about 20,000 to 50,000 yrs old. Illustrative Example A piece of charcoal from the ruins of a settlement in Japan was found to have 14C/12C ratio that was 0.617 times that formed in living organism. How old is this piece of charcoal. Given the half life of 14C is 5,770 years. [log 1.62 = 0.210] Solution We know 0.693 2.303 N log o t1/ 2 t Nt 0.693 2.303 1 log 5770 t 0.617 t = 4027 years Application: Tracers Radioisotopes are used as tracers to find the reaction mechanism. For example O O C6H5 C * + CH3OH C6H5 C + H2O OH * OCH3 Radioisotopes in Medicine • 1 out of every 3 hospital patients will undergo a nuclear medicine procedure • 24Na, t½ = 14.8 hr, emitter, blood-flow tracer • 131I, t½ = 14.8 hr, emitter, thyroid gland activity • 123I, t½ = 13.3 hr, -ray emitter, brain imaging Brain images with 123I-labeled compound 23.7 Illustrative Example The half life of W-238, which decays to Pb-206 is 4.5 x 109 years. A rock containing equal numbers of atoms of two isotopes would be how much old? Solution W238 Pb206 0.693 t1 / 2 4.5 109 yrs. k 2.303 N k log o t Nt No Initial no. of atoms of W238 Nt No. of atoms of W238 after certain time t Age of the rock 0.693 2.303 2N log t 4.5 109 t Nt 2.303 t 4.5 109 log2 4.5 109 yrs. 0.693 Illustrative Example A small amount of solution containing Na24 radioisotope with activity 2 × 103 dps was injected into blood of a patient. After 5 hr, a sample of blood drawn out from the patient showed an activity of 15 dpm ml–1 half-life of Na24 is 16 hr. Find the volume of blood in the patient. Solution Let the volume of blood be V ml. Activity of Na24 in V ml blood = 2 ×103 dps = 2 × 60 × 103 dpm Activity of Na24 in V ml blood after 5 hr = 15 × V dpm 0.693 0.693 0.0462 hr 1 t1/ 2 16 Now 0.693 N t log 0 N 0.693 120 103 or 5 log 0.0462 15 V V 3.713 103 Thank you