1 / 42

½ - life

½ - life. The decay of a single nuclei is totally random However, with large numbers of atoms a pattern does occur. Number of nuclei undecayed. time. half-life (t ½ ). ½ - life. This is the time it takes half the nuclei to decay. ½ - life.

carson-glass
Download Presentation

½ - life

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. ½ - life • The decay of a single nuclei is totally random • However, with large numbers of atoms a pattern does occur

  2. Number of nuclei undecayed time half-life (t½) ½ - life • This is the time it takes half the nuclei to decay

  3. ½ - life • This is the time it takes half the nuclei to decay Number of nuclei undecayed time half-life (t½)

  4. ½ - life • This is the time it takes half the nuclei to decay Number of nuclei undecayed A graph of the count rate against time will be the same shape time half-life (t½)

  5. Different ½ - lives • Different isotopes have different half-lives • The ½-life could be a few milliseconds or 5000 million years! Number of nuclei undecayed time half-life (t½)

  6. Example

  7. Example • A radio-isotope has an activity of 400 Bq and a half-life of 8 days. After 32 days the activity of the sample is • A – 200 Bq • B – 100 Bq • C – 50 Bq • D – 25 Bq

  8. Example • Asamplecontainsanamountofradioactivematerialwith ahalf-lifeof3.5days.After 2weeksthefraction oftheradioactivematerialremainingis • A. 94%. • B. 25%. • C. 6%. • D. 0%.

  9. Example • Nuclide Xhasahalf-lifeof1dayandnuclideYhasahalf-lifeof5 days.Inaparticularsample, theactivitiesof XandYarefoundtobeequal.Whenthe activityistestedagainafter10 days,the activitywill be • A. entirelyduetonuclideX. • B. dueequallytonuclidesXandY. • C. mostlyduetonuclideX. • D. mostlydue tonuclideY.

  10. Nuclear Reactions

  11. Transmutation • changing a nucleus by adding nucleons.

  12. Fusion • is the process by which two or more atomic nuclei join together, or "fuse", to form a single heavier nucleus.

  13. Fission • either a nuclear reaction or a radioactive decay process in which the nucleus of an atom splits into smaller parts (lighter nuclei).

  14. Unified mass unit (u) • Defined as 1/12 of the mass of an atom of Carbon-12 u = 1.6605402 x 10-27 kg

  15. Energy mass equivalence • E = mc2 • E = 1.6605402 x 10-27 x (2.9979 x 108)2 • E = 1.4923946316 x 10-10 J • Remembering 1 eV = 1.602177 x 10-19 J • 1 u = 931.5 MeV

  16. Mass defect For helium, the mass of the nucleus = 4.00156 u But, the mass of two protons and two nuetrons = 4.0320 u!!!! Where is the missing mass?

  17. Mass defect The missing mass (mass defect) has been stored as energy in the nucleus. It is called the binding energy of the nucleus. It can be found from E = mc2

  18. Mass defect calculation • Find the mass defect of the nucleus of gold, 196.97 - Au

  19. Mass defect calculation • The mass of this isotope is 196.97u • Since it has 79 electrons its nuclear mass is 196.97u – 79x0.000549u = 196.924u

  20. Mass defect calculation • The mass of this isotope is 196.97u • Since it has 79 electrons its nuclear mass is 196.97u – 79x0.000549u = 196.924u • This nucleus has 79 protons and 118 neutrons, individually these have a mass of 79x1.0007276u + 118x1.008665u = 198.080u

  21. Mass defect calculation • The mass of this isotope is 196.97u • Since it has 79 electrons its nuclear mass is 196.97u – 79x0.000549u = 196.924u • This nucleus has 79 protons and 118 neutrons, individually these have a mass of 79x1.0007276u + 118x1.008665u = 198.080u • The difference in mass (mass defect) is therefore 1.156u

  22. Mass defect calculation • The difference in mass (mass defect) is therefore 1.156u • This “missing mass” is stored as energy in the nucleus (binding energy). • 1u is equivalent to 931.5 MeV

  23. Binding energy This is the work required to completely separate the nucleons of the nucleus.

  24. Binding energy per nucleon This is the work required to completely separate the nucleons of the nucleus divided by the number of nucleons. It is a measure of how stable the nucleus is.

  25. The binding energy curve

  26. Example

  27. Let’s do some reading! • Page 381(380)to 385

  28. Nuclear Fission

  29. Uranium Uranium 235 has a large unstable nucleus.

  30. Capture A lone neutron hitting the nucleus can be captured by the nucleus, forming Uranium 236.

  31. Capture A lone neutron hitting the nucleus can be captured by the nucleus, forming Uranium 236.

  32. Fission The Uranium 236 is very unstable and splits into two smaller nuclei (this is called nuclear fission)

  33. Free neutrons As well as the two smaller nuclei (called daughternuclei), two neutrons are released (with lots of kinetic energy)

  34. Fission These free neutrons can strike more uranium nuclei, causing them to split.

  35. Chain Reaction If there is enough uranium (critical mass) a chain reaction occurs. Huge amounts of energy are released very quickly.

  36. Chain Reaction If there is enough uranium (critical mass) a chain reaction occurs. Huge amounts of energy are released very quickly.

  37. Bang! This can result in a nuclear explosion!YouTube - nuclear bomb 4

  38. Nuclear fusion – Star power!

  39. The binding energy curve

  40. Questions! • Page 379 Questions 3, 4, 8, 9, 11, 12, • Page 387 Questions 1, 6, 9, 10.

More Related