3. Atomic Structure The current model of atomic structure was not understood until the 20 th century. ▪ 1808 - John Dalton – new idea - the matter is made of atoms (tiny indivisible spheres) ▪ 1897 – J. J. Thomson discovered that all matter contains tiny negatively ‑ charged particles. → Thomson’s “plum-pudding” model – the atom was a positive sphere of matter and the negative electrons were embedded in it electrons (the plums) positive matter (pudding) He showed that these particles are smaller than an atom. He actually found the first subatomic particle ‑ the electron. Describing the emission and absorption spectrum of common gases

4. ▪ Ernest Rutherford got his students Geiger and Marsden to fire the fast moving α ‑ particles at very thin gold foil and observe how they were scattered. Rutherford Rutherford source of α: radioactive radon 1919 – Ernest Rutherford discovered proton ■ Scientists then set out to find the structure of the atom. so alpha particles only travel through a thin layer of atoms. Describing the emission and absorption spectrum of common gases

5. Expected result of Rutherford's experiment if the "plum pudding" model of the atom was correct. Actual result: Most of the α ‑ particles passed straight through the foil, some are slightly deflected, as expected, but to his surprise a few were scattered back towards the source. Rutherford said that this was rather like firing a gun at tissue paper and finding that some bullets bounce back towards you! Alpha particles would travel more or less straight through the atom without deflection.

6. Rutherford’s conclusion (1911): ▪ a particle had a head-on collision with a heavier particle ▪ heavier particle had to be very small, since very few a particles were bounced back. ▪ heavy particles must be positive (repulsion) Rutherford proposed that the positive charge of the atom was located in the center, and he coined the term nucleus. Atom contains a small but very massive positive nucleus, surrounded by negatively charged electrons at relatively large distances from it. The most suprising thing about this model is that the atom is mainly empty space! Using this model Rutherford calculated that the diameter of the gold nucleus could not be larger than 10 -14 m. → nuclear (planetary) model of the atom:

7. Gold nucleus (+79e) r Loss in kinetic energy must be equal to gain in potential energy of α particle in the field of gold nucleus. Rutherford used an a source given to him by Madame Curie. The inital a energy was ~ 7.7MeV (KE initial = 7.7 x 10 6 x 1.6 x 10 -19 J = 12 x 10 -13 J). α paricle is brought momentarily to rest (“having climbed as far as it can up the electrostatic hill”) when changing direction of the motion. The speed and hence the kinetic energy is zero, all the energy is now electrostatic potential energy. IDEA: Calculate the distance of an alpha particle’s closest approach to a gold nucleus. r ~ 3x10 -14 m Radius of gold atoms is ~ 3 ×10 -10 m. So a nucleus is at least 10 000 times smaller than an atom. It is important to emphasise that this calculation gives an upper limit on the size of the gold nucleus; we cannot say that the alpha particle touches the nucleus; a more energetic a might get closer still.

8. no neutrons (AD 1911) ▪ Electrons are orbiting around nucleus in circles. (if they were not orbiting but at rest, they would move straight to the nucleus; instead centripetal force is provided by the electrostatic attraction between electrons and nucleus.) Problems with Rutherford’s model: ▪ According to Maxwell, any accelerating charge will generate an EM wave ▪ Electrons will radiate, slow down and eventually spiral in to nucleus. The end of the world as we know it. ▪ The solution was found in quantum theory. Nuclear Atom Rutherford/Planetary/Nuclear Model Rutherford/Planetary/Nuclear Model of atom: the atom consists of a very tiny but very massive positive nucleus, surrounded by electrons that orbit the nucleus as result of electrostatic attraction between the electrons and the nucleus.

9. ▪ Electrons could only exist in certain orbits at specific energy levels (“discrete states”), called “stationary states.” ▪ n is called the principal quantum number and goes all the way up to infinity (  )! ▪ Electrons in these stationary states do not emit EM waves as they orbit. Bohr’s Model - atomic energy levels Thinkquest http://library.thinkquest.org/16468/gather/english.htm http://library.thinkquest.org/16468/gather/english.htm ▪ If a photon of just the right energy strikes an atom, it is absorbed by the atom causing an electron to jump to a higher energy level: ▪ We say the atom is excited ▪ Electron stays in higher energy level only for about 1 ns. ▪ When the atom de-excites the electron jumps back to a lower energy level emitted/absorbed photon has:

10. Transitions between energy levels excitation de-excitation These energies naturally lead to the explanation of the hydrogen atom spectrum: ▪ In its ground state or unexcited state, hydrogen’s single electron is in the 1 st energy level (n = 1): Albert Einstein: The frequency of emitted light (photon) is proportional to the change of energy of the electron: Energy of that photon is equal to the energy difference between two states. E photon = ΔE

11. PRACTICE: An electron jumps from energy level n = 3 to energy level n = 2 in the hydrogen atom. (a) What series is this de-excitation in? Transitions between energy levels (b) Find the atom’s change in energy in eV and in J. ▪ The hydrogen atom lost 3.02  10 -19 J of energy. ▪ From conservation of energy a photon was created having E = 3.02  10 -19 J. ▪ E = hf  f = E/h = 3.02  10 -19 / 6.63  10 -34 f = 4.56  10 14 Hz. ▪ c = f  = c/f = 3.00  10 8 /4.56  10 14 = 6.58  10 -7 m. ▪ = 658  10 -9 m = 658 nm. ▪ When finding f, be sure E is in Joules, not eV. ▪ Find it on the diagram: ▪ This jump is contained in the Balmer Series, and produces a visible photon ▪  E = E f – E 0 = –3.40 –(–1.51) = – 1.89 eV. ▪  E = (– 1.89)(1.60  10 -19 )J = – 3.02  10 -19 J. (c) Find the energy (in J) of the emitted photon. (d) Find the frequency of the emitted photon. (e) Find the wavelength (in nm) of the emitted photon.

12. Discrete energy and discrete energy levels ▪ Discrete means discontinuous, or separated PRACTICE: Which one of the following provides direct evidence for the existence of discrete energy levels in an atom? A. The continuous spectrum of the light emitted by a white hot metal. B. The line emission spectrum of a gas at low pressure. C. The emission of gamma radiation from radioactive atoms. D. The ionization of gas atoms when bombarded by alpha particles.

13. PRACTICE: A spectroscopic examination of glowing hydrogen shows the presence of a 434 nm blue emission line. (a) What is its frequency? (b) What is the energy (in J and eV) of each of its blue-light photons? ▪ Because it is visible use the Balmer Series with ∆E = -2.86 eV. ▪ Note that E 2 – E 5 = – 3.40 – (–0.544) = –2.86 eV. ▪ Thus the electron jumped from n = 5 to n = 2. ▪ c = f  f = c/ = 3.00  10 8 /434  10 -9 ▪ f = 6.91  10 14 Hz ▪ E = hf = (6.63  10 -34 )(6.91  10 14 ) E = 4.58  10 -19 J. ▪ E = (4.58  10 -19 J)(1 eV/ 4.58  10 -19 J) E = 2.86 eV. (c) What are the energy levels associated with this photon?

14. PRACTICE: The element helium was first identified by the absorption spectrum of the sun. continuous spectrum absorption spectrum emission spectrum One of the wavelengths of the absorption spectrum for helium occurs at 588 nm. (b) Show that the energy of a photon having a wavelength of 588 nm is 3.38  10 -19 J. ▪ An absorption spectrum is produced when a cool gas is between a source having a continuous spectrum and an observer with a spectroscope. ▪ The cool gases absorb their signature wavelengths from the continuous spectrum. ▪ Where the wavelengths have been absorbed by the gas there will be black lines. (a) Explain what is meant by the term absorption spectrum. The diagram represents some energy levels of the helium atom. (c) Use the information in the diagram to explain how absorption at 588 nm arises.

15. And it was. Soon enough. Schrodinger: The modern model – wave equations probability wave equations The model we now accept is that there is a nucleus at the centre of the atom and the electrons do exist in certain energy levels, but they don’t simply orbit the nucleus. The probability of finding electron somewhere is given by wave equations, resulting in some interesting patterns. The result of this theory can be again visualized using very simple model, this time only energy level model. This model is not a picture of the atom but just represents possible energy of electrons. only for hydrogen hydrogen like ions, Bohr's model was so successful that he immediately received world-wide fame. Unfortunately, Bohr's model worked only for hydrogen. It was tried to apply to hydrogen like ions, but was not so successful. Thus the final atomic model was yet to be developed.

16. When a gas in a tube is subjected to a voltage, the gas ionizes, and in process of recombination it emits light. Describing the emission and absorption spectrum of common gases

17. We can analyze that light by looking at it through a spectroscope. A spectroscope acts the same way as prism, in that it separates the incident light into its constituent wavelengths. For example, heated barium gas will produce an emission spectrum that looks like this: An emission spectrum is an element fingerprint. 400 450 500 550 600 650 700 750 /  10 -9 m ( / nm)

18. light source cool gas continuous spectrum absorption spectrum emission spectrum compare… Same fingerprint! hot gas continuous spectrum: ▪ A hot solid, liquid or gas at high pressure produces a continuous spectrum: all λ. emission-line spectrum: ▪ A hot, low-density / low pressure gas produces an emission-line spectrum: only specific λ. ▪ A continuous spectrum source viewed through a cool, low-density gas absorption-line spectrum produces an absorption-line spectrum – missing λ – dark lines. Thus when we see a spectrum we can tell what source we are seeing.

19. ● Each element (atom/ion) produces a specific set of absorption (and emission) lines. We call this the "spectral signature" or “fingerprints” of an atom/ion. ● Allows the identification of elements across the galaxy and universe. (If we mapped it and can recognize it) Emission Spectra Absorption Spectra Thus when we see a spectrum we can tell what source we are seeing.

20. ▪ In a later lecture we will discover that the most intense light reaching us from the sun is between 500 nm and 650 nm in wavelength. ▪ Evolutionarily our eyes have developed in such a way that they are most sensitive to that range of wavelengths, as shown in the following graphic:

21. Nuclear Structure

22.  Recall the mass spectrometer, in which an atom is stripped of its electrons and accelerated through a voltage into a magnetic field.  Scientists discovered that hydrogen nuclei had three different masses:  Since the charge of the hydrogen nucleus is e, scientists postulated the existence of a neutral particle called the neutron, which added mass without charge. Isotopes For the element hydrogen, it was found that its nucleus existed in three forms: Hydrogen Deuterium Tritium Isotopes ▪ An element’s chemistry is determined by the number of protons/electrons surrounding it, therefore isotopes of an element have the same chemical properties. Water, H 2 O and heavy water, D 2 O have exactly the same chemical properties. ▪ But heavy water is slightly denser than water.

23. ● Nucleon ● Nucleon The name given to the particles of the nucleus. ● Nuclide or Species ● Nuclide or Species A particular combination of protons and neutrons that form a nucleus. It is used to distinguish isotopes among nuclei. ● Nucleon number (mass number) - A The number of protons plus neutrons in the nucleus. ● Proton number - Z The number of protons in the nucleus. ●Isotopes ● Isotopes Nuclei (atoms) with the same number of protons but different numbers of neutrons. ● Neutron number - N ( ● Neutron number - N (N = A – Z) The number of neutrons in the nucleus. ● Symbol for a nuclide

24. Isotopes – Nuclei (atoms) of the same element differing in masses due to different numbers of neutrons (the same proton number, different nucleon number). because there is no other way to explain the mass difference of two isotopes of the same element. ▪ The existence of isotopes is evidence for the existence of neutrons, because there is no other way to explain the mass difference of two isotopes of the same element. ▪ The same number of electrons – the same bonding - the same chemical properties ▪ Different masses – different physical properties ▪ Many isotopes do not occur naturally, and the most massive isotope found in nature is uranium isotope ▪ About 339 nuclides occur naturally on Earth, of which 269 (about 79%) are stable ▪ the current largest atomic number element, with atomic number 118, survived for less than a thousandth of a second click me!

25. PRACTICE: Which of the following gives the correct number of electrons, protons and neutrons in the neutral atom 65 29 Cu?  A = 65, Z = 29, so N = A – Z = 65 – 29 = 36.  Since it is neutral, the number of electrons equals the number of protons = Z = 29. Isotopes

26. PRACTICE: Ag-102, Ag-103 and Ag-104 are all isotopes of the element silver. Which one of the following is a true statement about the nuclei of these isotopes? A. All have the same mass. B. All have the same number of nucleons. C. All have the same number of neutrons. D. All have the same number of protons. Isotopes of an element have different masses and nucleon totals.  Isotopes of an element have the same number of protons, and by extension, electrons. This is why their chemical properties are identical. Isotopes

27. PRACTICE: Track X shows the deflection of a singly-charged carbon-12 ion in the deflection chamber of a mass spectrometer. Which path best shows the deflection of a singly- charged carbon-14 ion? Assume both ions travel at the same speed. SOLUTION:  Since carbon-14 is heavier, it will have a bigger radius than carbon-12.  Since its mass is NOT twice the mass of carbon-12, it will NOT have twice the radius. Isotopes

28. The strong nuclear force

29. Protons are positive, neutrons are neutral (a nucleus is roughly 10 -15 m, they would drift apart if put them together), so if the electric force was the only force involved, you couldn’t create a nucleus. There has to be some other force that holds protons and neutrons together and it must be stronger than the electric force. Well, in a brilliant stroke of imagination, physicists have named this force "the strong force." Althoug the nuclear force is strong, nuclei do not attract each other, so that force must be very short range, unlike the electric force that extends forever. In a nutshell, the strong force… (1) counters the Coulomb force to prevent nuclear decay, therefore must be very strong. (2) is very short-range, since protons located far enough apart do, indeed, repel. The strong nuclear force was first described by the Japanese physicist Hideki Yukawa in 1935. It is the strongest force in the universe, 10 38 times stronger than gravitational force and 100 times stronger than the electromagnetic force. What holds the nucleus together?

30. It is the force which attracts protons to protons, neutrons to neutrons, and protons and neutrons to each other. That force has a very short range, about 1.5 radii of a proton or neutron (1.5 x 10 -14 m) and is independent of charge and this is the reason the nucleus of an atom turns out to be so small. If the protons can't get that close, the strong force is too weak to make them stick together, and competing electromagnetic force can influence the particles to move apart. As long as the attractive nuclear forces between all nucleons win over the repulsive Coulomb forces between the protons the nucleus is stable. It happens as long as the number of protons is not too high. Atomic nuclei are stable subject to the condition that they contain an adequate number of neutrons, in order to "dilute" the concentration of positive charges brought about by the protons. The most massive isotope found in nature is uranium isotope For more massive nuclei strong nuclear force can’t overcome electric repulsion.

31. Fundamental forces and their properties GRAVITY STRONG ELECTROMAGNETICWEAK + + nuclear force light, heat and charge radioactivity freefall ELECTRO-WEAK WEAKEST STRONGEST Range: Extremely Short Range:  Range: Short Range:  Force Carrier: Gluon Force Carrier: Photon Force Carrier: Graviton

32. Fundamental forces and their properties PRACTICE: The nucleus of an atom contains protons. The protons are prevented from flying apart by A. The presence of orbiting electrons. B. The presence of gravitational forces. C. The presence of strong attractive nuclear forces. D. The absence of Coulomb repulsive forces at nuclear distances.  It is the presence of the strong force within the nucleus. PRACTICE: Use Coulomb’s law to find the repulsive force between two protons in a helium nucleus. Assume the nucleus is 1.00  10 -15 m in diameter and that the protons are as far apart as they can get.  F = ke 2 / r 2 = 9  10 9 (1.6  10 -19 ) 2 / (1.00  10 -15 ) 2  F = 230 N.  From chemistry we know that atoms can be separated from each other and moved easily.  This tells us that at the range of about 10 -10 m (the atomic diameter), the strong force is zero.

33. Radioactivity

34. In 1893, Pierre and Marie Curie announced the discovery of two radioactive elements, radium and polonium. When these elements were placed by a radio receiver, that receiver picked up some sort of activity coming from the elements Studies showed this radioactivity was not affected by normal physical and chemical processes. Radioactive decay In 1896, while studying a uranium compound, French scientist Henri Becquerel discovered that a nearby photographic plate had somehow been exposed to some source of "light" even though it had not been uncovered Apparently the darkening of the film was caused by some new type of radiation being emitted by the uranium compound. This radiation had sufficient energy to pass through the cardboard storage box and the glass of the photographic plates.

35. HyperPhysics ● Unstable nucleus by emitting radioactive particle (energy) becomes more stable. Radioactity Radioactivity refers to the particles which are emitted from nuclei as a result of nuclear instability. Because the nucleus experiences the intense conflict between the two strongest forces in nature, it should not be surprising that there are many nuclear isotopes which are unstable and emit some kind of radiation. ● The Most common types of radiation are called alpha (  ), beta (  ), and gamma (  ) radiation. ● But there are several other varieties of radioactive decay.other varieties Alpha particles, beta particles, and gamma rays are NOT radioactive themselves

36. Alpha radiation (decay) ▪ Conservation of mass/energy (total energy) Mass of parent > mass of daughter + mass of alpha Difference = kinetic energy ▪ α decay occurs primarily among heavy elements: nucleus has too many protons which cause excessive repulsion. In an attempt to reduce the repulsion, a helium nucleus is emitted. ▪ All alpha particles consistently have an energy of about 5 MeV ▪ Since the energy needed to knock electrons off of atoms is just about 10 eV, one alpha particle can ionize a lot of atoms ▪ It is just this ionization process that harms living tissue, and is much like burning at the cell level

37. Beta radiation (decay) ▪ This means that the proton number increases by 1, while the total nucleon number remains the same. electrons/positrons ▪ Beta particles are high energy electrons/positrons emitted from the nucleus. antineutrino 14 C  14 N + + e - ▪ In  + decay, a proton becomes a neutron and a positron is emitted from the nucleus. 10 C  10 B + + e + (nu) ▪ This means that the proton number decreases by 1, while the total nucleon number remains the same. In contrast to the alpha particle, it was discovered that beta particles could have a large variety of kinetic energies.

38. Antineutrino produced in beta decay is antiparticle of neutrino. It was in radioactive beta decay that the existence of the weak interaction was first revealed. Weak interaction causes the transmutation p → n. Neutrino was introduced into theory in 1930 due to the fact that energy was missing in observations of beta decay. Pauli theorized that an undetected particle was carrying away the observed difference between the energy. Quite a number of antineutrinos are discovered streaming from the planet's centre produced by natural radioactivity in the Earth responsible for the immense quantity of heat generated by Earth. Neutrinos are elementary particles that travel close to the speed of light, lack an electric charge, are able to pass through ordinary matter almost undisturbed and are thus extremely difficult to detect It took 26 years before the neutrino was actually detected. As of 1999, it is believed neutrinos have a minuscule, but nonzero mass. They are usually denoted by the Greek letter ν (nu) Neutrinos are created as a result of “beta plus” decay in which proton is converted via weak force to a neutron, a positron (antielectron) and a neutrino (nuclear fusion powering the sun and other stars.). Most neutrinos passing through the Earth emanate from the sun, and more than 50 trillion solar electron neutrinos pass through the human body every second. first two steps in the sun

39. Gamma radiation (decay) Gamma-emission does not change the structure of the nucleus, but it does make the nucleus more stable because it reduces the energy of the nucleus. Nuclear energy levels are of the order of MeV hence the high energy of the emitted photon, and the frequencies (f = E/h) correspond to gamma rays. Recall that electrons in an atom moving from an excited state to a de-excited state release a photon Nuclei can also have excited states. When a nucleus de-excites, it also releases a photon. This process is called gamma (  ) decay. In α and β decay, the product of decay is often nuclide in an excited state. The daughter nuclide then drops to its ground state by emitting a photon. 234 Pu*  234 Pu + 

40. Decay chains A radioactive nuclide often produces a radioactive daughter nuclide. The daughter will also decay, and the process will continue until finally a stable nuclide is formed. This process is known as decay chain.

41. Properties Properties The diagram shows how the different types are affected by a magnetic field. ▪The alpha beam is a flow of positively (+) charged particles, so it is equivalent to an electric current. It is deflected in a direction given by right ‑ hand rule ‑ the rule used for working out the direction of the force on a current ‑ carrying wire in a magnetic field. ▪ Beta particles are much lighter than the alpha particles and have a negative charge, so they are deflected more, and in the opposite direction. ▪ Being uncharged, the gamma rays are not deflected by the field.

42. Ionising Properties ● Radiation ionises molecules by `knocking' electrons off of them. ● As it does so, energy is transferred from the radiation to the material. ● To knock an electron out of an atom requires about 10 eV Since the α-particle is massive, relatively slow-moving particle (up to 0.1 c) with a charge of +2e, it interacts strongly with matter. Alpha particles have energies of about 5 MeV so α-particle can ionize a lot of atoms before they loose all their KE, passing through just a few cm of air They cannot penetrate paper. Can be very harmful since ionizing atoms of human tissue cause demage to the cells similar to burning. α-particle

43. β-particle Neutrinos can go through miles of lead!

44. Alpha, Beta and Gamma Properties

45. Protection against radiation There are two ways we can reduce the affect of nuclear radiation: distance and sheilding. Alpha and beta radiation have a very short range in air, so will not be dangerous a few meters away from the source. The number of gama photons decreases proportional to 1/r 2 (where r is the distance from the source), so the further away you are, the safer you will be. Although alpha is the most ionizing radiation, it can be stopped by a sheet of paper (although that means that alpha is the most harmful if ingested). Beta and gamma are more penetrating, so need a thick lead shield to provide protection. A radiation burn caused during radiotherapy for cancer

46. Average annual human exposure to ionizing radiation in millisieverts (mSv) Natural radiation sourceWorldUSAJapanRemark Inhalation of air1.262.280.4 mainly from radon, depends on indoor accumulation Ingestion of food & water0.290.280.4K-40, C-14, etc. Terrestrial radiation from ground0.480.210.4 depends on soil and building material. Cosmic radiation from space0.390.330.3depends on altitude sub total (natural)2.43.11.5 Background radiation  Background radiation is the ionizing radiation that people are exposed to in everyday life, including natural and artificial sources

47. Average annual human exposure to ionizing radiation in millisieverts (mSv) Artificial radiation sourceWorldUSAJapanRemark Medical0.632.3CT scans excludes radiotherapy Consumer items-0.13 cigarettes, air travel, building materials, etc. Atmospheric nuclear testing0.005-0.01 peak of 0.11 mSv in 1963 and declining since Occupational exposure0.005 0.01 radon in mines, medical and aviation workers Nuclear fuel cycle0.00020.001 up to 0.02 mSv near sites; excludes occupational Other-0.003 Industrial, security, medical, educational, and research sub total (artificial)0.613.142.33 Background radiation

48. Stability If you plot the neutron number N against the proton number Z for all the known nuclides, you get the diagram shown here

49. ● It is the strong nuclear force that holds the nucleons together, but this is a very short range force. ● The repulsive electric force between the protons is a longer range force. ● So in a large nucleus all the protons repel each other, but each nucleon attracts only its nearest neighbours. More neutrons are needed to hold the nucleus together (although adding too many neutrons can also cause instability). There is an upper limit to the size of a stable nucleus; all the nuclides with Z higher than 83 are unstable. The stable nuclides of the lighter elements have approximately equal numbers of protons and neutrons? However, as Z increases the `stability line' curves upwards. Heavier nuclei need more and more neutrons to be stable. Can we explain why?

50. Half - life

51. ● Suppose you have a sample of certain number of identical unstable nuclei. ● All the nuclei are equally likely to decay, but you can never predict which individual nucleus will be the next to decay. ● The decay process is completely random. ● Also, there is nothing you can do to `persuade' one nucleus to decay at a certain time. ● The decay process is spontaneous. Does this mean that we can never know the rate of decay? No, because for any particular radioactive nuclide there is a certain probability that an individual nuclide will decay. This means that if we start with a large number of identical nucleides we can predict how many will decay in a certain time interval. rate of decay ∞ number of nuclei

52. Let’s say you have two identical atoms. One atom decayed and the other did not. Why? Why would one wait and the other one not ? What is the difference between them? The answer to this is not easy. Quantum mechanics and probability must be called upon. Einstein never accepted Quantum Mechanics, and this part of the theory (probability that something would happen and random processes). He summarised his objections by saying "God does not play at dice with the universe." Bohr responded "Quit telling God what to do!" Einstein rejected the idea that completely identical initial states can evolve to different outcomes. They must be different in the first place. SOMETHING DIFFERENT!!!

53. Definition H alf-life (T 1/2 ) is the time taken for one half of the nuclei present in any given radioactive sample to decay. time Number of nuclei remaining t ½ t ½ t ½ t ½ N0N0 2 N0N0 4 N0N0 8 N0N0 …....

54.  Rather than measuring the amount of remaining radioactive nuclide there is in a sample (which is extremely hard to do) we measure instead the decay rate (which is much easier).  Decay rates are measured using various devices, most commonly the Geiger- Mueller counter.  Decay rates are measured in Becquerels (Bq). 1 Bq  1 decay / second Becquerel definition Activity and half-life

55. ● It is much easier to measure the radiation than number of undecayed nuclei in a sample. ● Activity (becquerel - Bq)of a radioactive sample is the average number of disintegrations per second. ● 100 Bq means that 100 nuclei are disintegrating/sec. time / days activity / Bq 40 20 10 5 0 0 8 16 24 half-life original activity = 40 counts/sec Activity of a sample of I -131. T 1/2 = 8 days As the activity is always proportional to the number of undecayed nuclei, it too halves every 8 days. Since the rate of decay is proportional to the number of nuclei, a graph of the rate of particle emission against time will have the same shape.

56. time activity Radioactive decay is a random process. So, in practice, the curve is a ‘best fit’ of points which vary irregularly like this.

57. Definition 2 The half-life of a radioactive isotope is the time taken for the activity of any given sample to fall to half its original value. Exponential Decay Any quantity that reduces by the same fraction in the same period of time will follow an exponential decay curve. The half life can be calculated from decay curves Take several values and the take an average

58. sample containing N radioactive atoms, grams, kilogram, moles,… T 1/2 after T 1/2 N/2 decayed T 1/2 after T 1/2 N/2/2 decayed T 1/2 after T 1/2 N/2/2/2 decayed … transmutated Example: T 1/2 after time n T 1/2 only survived T 1/2 for example, after 4T 1/2 there is still atoms in the sample (survived) and transmutated

59. Example: Cobalt–60 decays by beta emission and has a half-life of aproximately 5 years. If a sample of cobalt–60 emits 40 beta particles per second, how many will the same sample be emitting in 15 years time? After 5 years activity will be 20/sec (number of decays/sec). After another 5 years it will be 10/sec. Finally after a further 5 years it will emit 5 particles/sec.

61. EXAMPLE: Suppose you have 64 grams of a radioactive material which decays into 1 gram of radioactive material in 10 hours. What is the half-life of this material?  The easiest way to solve this problem is to keep cutting the original amount in half... 64 t half 32 t half 16 t half 8 4 2 1  Note that there are 6 half-lives in 10 h = 600 min. Thus t half = 100 min. EXAMPLE: A nuclide X has a half-life of 10 s. On decay a stable nuclide Y is formed. Initially, a sample contains only the nuclide X. After what time will 87.5% of the sample have decayed into Y? A. 9.0 s B. 30 s C. 80 s D. 90 s  We want only 12.5% of X to remain.  Thus t = 3t half = 3(10) = 30 s. 100% t half 50% t half 25% t half 12.5%

62. PRACTICE: A sample of radioactive carbon-14 decays into a stable isotope of nitrogen. As the carbon-14 decays, the rate at which nitrogen is produced A. decreases linearly with time. B. increases linearly with time. C. decreases exponentially with time. D. increases exponentially with time.  The key here is that the total sample mass remains constant. The nuclides are just changing in their proportions. Carbon Nitrogen  Note that the slope (rate) of the red graph is decreasing exponentially with time. PRACTICE: An isotope of radium has a half-life of 4 days. A freshly prepared sample of this isotope contains N atoms. The time taken for 7N/8 of the atoms of this isotope to decay is A. 32 days. B. 16 days. C. 12 days. D. 8 days. If 7N / 8 has decayed, only 1N / 8 atoms of the isotope remain.  N  (1/2)N  (1/4)N  (1/8)N is 3 half-lives.  That would be 12 days since each half-life is 4 days.

63. PRACTICE: Radioactive decay is a random process. This means that A. a radioactive sample will decay continuously. B. some nuclei will decay faster than others. C. it cannot be predicted how much energy will be released. D. it cannot be predicted when a particular nucleus will decay. Random process PRACTICE:  Isotopes of an element have the same number of protons and electrons, but different numbers of neutrons

64.  The lower left number in the symbol is the number of protons.  Since protons are positive, the new atom has one more positive value than the old.  Thus a neutron decayed into a proton and an electron (  - ) decay.  And the number of nucleons remains the same… -- 42 Flip the original curve so the amounts always total N 0

65.  It is the gamma decay that leads us to the conclusion that excited nuclei, just like excited atoms, release photons of discrete energy, implying discrete energy levels.  -ray decay happens when the nucleus goes from an excited state to a de-excited state.  Since the ratio is 1/2, for each nickel atom there are 2 cobalt atoms.  Thus, out of every three atoms, 1 is nickel and 2 are cobalt.  Therefore, the remaining cobalt is (2/3)N 0.

66. N doubled, so A did too. Activity is proportional to the number radioactive atoms.  But the half-life is the same for any amount of the atoms…

67. Activity is proportional to the number radioactive atoms remaining in the sample.  Since X’s half-life is shorter than Y’s, less activity will be due to X, and more to Y at any later date…  60 days is 2 half-lives for P so N P is 1/4 of what it started out as.  60 days is 3 half-lives for Q so N Q is 1/8 of what it started out as.  Thus N P / N Q = (1/4) / (1/8) = (1/4)  (8/1) = 8/4 = 2.