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Chemistry PowerPoint Presentation

Chemistry

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Chemistry

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  1. Chemistry

  2. NUCLEAR CHEMISTRY

  3. Session opener Nuclear chemistry, sun & life

  4. Session objectives • Radioactivity • Nuclear reactions • Kinetics of radioactive decay • Radioactivity equilibrium • Fission reaction • Fusion reaction • Radiocarbon dating • Medical applications

  5. 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.

  6. X Y n/p too large beta decay n/p too small positron decay or electron capture 23.2

  7. This nucleus would be unstable and would tend to lower the ratio by emitting -particles and thus moves towards the zone of stability. 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

  8. 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

  9. Loss of –particle Loss of –particle Nuclear reactions

  10. Group displacement laws The emission of an a-particle results in the formation of an element which lies two places to the left and the emission of a b  -particle results in the formation of an element which lies one place to the right in the periodic table

  11. Illustrative Example is radioactive and emits a and b particles to form What is the respective values of a and b? Solution Let x number of a particles and y number of b particles get emitted. 238 = 4x + 206 or x = 8 92 = 2x – y + 82 or y = 6

  12. Solution Cont. 212Po 4He + AX 2 Z 84 212Po 4He + 208Pb 2 82 84 212Po decays by alpha emission. Write the balanced nuclear equation for the decay of 212Po. 212 = 4 + A A = 208 84 = 2 + Z Z = 82 23.1

  13. Name of the Series Initial element Stable element Value of n for initial element Value of n for the stable elementSeries 4n Thorium series Thorium–232 Lead–208 58 52 4n + 1 Neptunium series Neptunium–237 Bismuth–209 59 52 4n + 2 Uranium series Uranium–238 Lead–206 59 51 4n + 3 Actinium series Uranium–235 Lead–207 58 51 Disintegration series

  14. 4n series In the 4n series,all nuclides have mass number that are multiples of 4. 

  15. 4n +1 series In the 4n + 1 series,all nuclides have mass numbers that are one number greater than multiples of 4.

  16. 4n+2 series In the 4n + 2 series, all nuclides have mass number that are two numbers greater than multiples of 4.

  17. 4n+3 series In the 4n + 3 series, all nuclides have mass numbers that are three numbers greater than multiples of 4.

  18. Nuclear Binding Energy

  19. Nuclear binding energy BE = 0.1587 amu 1 amu = 1.49 x 10-10 J BE = 2.37 x 10-11J The energy required to break up a nucleus into its component protons and neutrons. 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

  20. Average Binding Energy as a Function of Atomic Number

  21. Kinetics of radioactive decay M D N0=Initial no. of radioactive particles N=No. of radioactive particles after time t

  22. The rate of disintegration ofM into D is equal to Kinetics of radioactive decay

  23. Average or Mean life Period Half life period Units 1 curie (C) = 3.7 × 1010 dps S.I unit is Beequerel (Bq) (Bq) = 1 disintegration per second.

  24. Calculation of decay constant Illustrative example Calculate the activity in terms of dpm of 0.001 g sample of Pu239 (half life = 24300 years). Solution:

  25. 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 Here a = 10–6 g  a – x = 6.09 × 10–7 g

  26. 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 N = 2.89 × 1015 6.023 × 1023 of Na24 atoms = 24 g

  27. 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 Number of a particles emit per second

  28. + 4He 17O 14N + → 1H 2 8 7 1 + 4He 30P 24Al + → 1n 2 15 13 0 15 14 +1 30P + 0 30Si → Shared Nobel Prize 1938 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. Irene Joliot-Curie.

  29. Cf U U U U 15 1 1 249 238 239 239 260 7 0 0 98 92 105 92 92 + n  + → + 0  → Np 239 4 n + + N → 93 -1 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.

  30. Fission of U-235

  31. Fission of U-235 The minimum amount of fissionable material required to continue a nuclear chain reaction is called critical mass.

  32. Nuclear Reactors

  33. n + 1 → 1 0 +  → 0 239U 238U 239Np 239U 239Np -1 239Pu 93 92 92 93 92 94 +  → 0 -1 Breeder Reactors A breeder reactor is one that produces more fissionable nuclei than it consumes i.e. it increases the concentration of fissionable nuclie.

  34. Fusion

  35. Difference between fission and fusion

  36. Rate law for reactions involving parallel reactions

  37. 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 On solving,

  38. Radioactive equilibrium 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.

  39. 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.

  40. 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 Here N = 1 g

  41. This carbon-14 is radioactive with a half life period of 5700 years. This is changed into nitrogen by the emission of Radio carbon dating The principle is based upon the formation of C14 by neutron capture in the upper atmosphere.

  42. Radio carbon dating The age of the object can be estimated. This method cannot be applied to estimate the age of an object which is about 20,000 to 50,000 yrs old.

  43. 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 \ t = 4027 years

  44. Application: Tracers Radioisotopes are used as tracers to find the reaction mechanism. For example

  45. Radioisotopes in Medicine • 1 out of every 3 hospital patients will undergo a nuclear medicine procedure • 24Na, t½ = 14.8 hr, b emitter, blood-flow tracer • 131I, t½ = 14.8 hr, b emitter, thyroid gland activity • 123I, t½ = 13.3 hr, g-ray emitter, brain imaging Brain images with 123I-labeled compound 23.7

  46. 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

  47. 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.

  48. 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 Now

  49. Thank you