Radioactivity
Many nuclides are unstable. There are three (common) types of radioactive decay: Alpha decay: a Helium nucleus is emitted Beta decay: an electron or positron is emitted Gamma decay: a photon is emitted We will consider the physics of each of these processes. 2
The probability for a given nuclide to decay is a constant. It does not change with time. It does not change if there are other nuclei nearby decaying as well. PRS Question: Suppose you start off with one nucleus. The probability for the nucleus to decay in time T is P. What is the probability to have a decay in time T if you have two nuclei? (assume P is small) A) 0 B) P C) 2P D) 4P E) none of the above 3
The correct answer is C) 2 times P. PRS Question: Suppose you start off with one nucleus. The probability for the nucleus to decay in time T is P. What is the probability to have a decay in time T if you have two nuclei? (assume P is small) A) 0 B) P C) 2P D) 4P E) none of the above 4
Therefore, the number of decays (dn) in a given length of time (dt) is proportional to N: The constant of proportionality is called the time constant λ (which has units of inverse time) dn dt dn dt / N = N minus sign because N is decreasing with increasing time We can solve this differential equation by integrating: Z N N 0 dn N = Z T 0 dt! N(T )=N 0 e T N0=N(0)=the initial number of nuclides. 5
The number of surviving nuclei follows an exponential decay with time. The number that decayed is N 0 N(T )=N 0 1 e T We can also define the half-live t1/2 as the time for half of the original sample of nuclei to decay: 1 2 N 0 = N 0 e therefore, ln 1 2 = t 1/2 t 1/2 so that t 1/2 = ln 2.693 6
Behavior of half-lives: At t=t1/2, N=N0/2 At t=2t1/2, N=N0/4 At t=3t1/2, N=N0/8 etc... Example: The half-life of U-235 is 7.2 10 8 years. What fraction will survive after 1 billion years (1 Gyr)? fraction surviving = N(T ) N 0 = N 0e T = ln 2 t 1/2 =9.72 10 10 y 1 N 0 = e T fraction surviving = e (9.72 10 10 ) (10 9) 37.8% 7
A Geiger counter measures the number of decays per second. This quantity is called the activity R. R dn dt = d dt N 0e T = N 0 e T = N(T ) Notice that the half-life t1/2 and decay constant λ are constant in time. But the activity decreases with time, and depends on the amount of the substance (i.e. how many nuclei there are). A common unit of radioactivity: 1 Becquerel (1 Bq) = 1 decay per second. 8
Carbon Dating Carbon in atmospheric CO2 is mostly Carbon-12. Cosmic rays strike the atmosphere, turning Nitrogen-14 into Carbon-14: 1 0n+ 14 7 N! 14 6 C+ 1 1H This happens at a steady rate. Plants absorb CO2, and animals ingest plans. So all living organism have the same ratio of C-12 to C-14. And they have the same activity, R=.255 Bq per gram of Carbon. When an organism dies, it stops absorbing CO2, so C-14 keeps decaying and the activity drops. Measuring the activity determines how long ago the organism died. 9
Example 1: A dead organism has activity today of 0.03 Bq per gram of Carbon. How long ago did it die? The half-life of carbon is 5740 years. = ln 2 t 1/2 = 0.693 5740 y =1.21 10 4 y 1 original activity: N 0 =0.255 Bq/g current activity: N = N 0 e T =0.03 Bq/g Solve for T: e T = 0.03 0.255 0.118 T = ln (1/0.118) 1.21 10 4 y 1 =1.77 105 y 10
Example 2: What fraction of Carbon in living organisms is C-14? From before, we know that the time constant is: =1.21 10 4 y 1 =3.82 10 12 s 1 And the activity is for a living organism is: R = N =0.255 Bq/g Solving for N gives: N = 0.255 Bq/g 3.82 10 12 s 1 =6.68 1010 atoms/g But Carbon is mostly C-12, so the number of carbon atoms per gram is N 12 =6.02 10 23 /12 = 5.02 10 22 atoms/g So the ratio is N = 6.68 1010 atoms/g N 12 5.02 10 22 atoms/g 1.3 10 12 11
γ-decay Just as the (hydrogen) atom can have excited states, the nucleus can also have excited states. When a nucleus in an excited state decays to a lower energy level, it will emit a photon (γ-ray): 12 6 C?! 12 6 C + All nuclei have many excited states. Another example: Cobalt-60 decays via beta decay to an excited Nickel state, which then decays to the ground-state Nickel and two photons: 60 27Co! e + 60 28Ni?! e + 60 28Ni + 2 12
α-decay A Helium nucleus a.k.a. an alpha particle, composed of 2 protons+2 neutrons, is emitted. Decay of parent into daughter: A ZP! A 4 Z 2 D+ 4 2He Energy conservation: m P c 2 = m D c 2 + K D + m c 2 + K The total net kinematic energy released is the disintegration energy or Q-value: Q K D + K =(m P m D m )c 2 13
Alpha decay will only occur if Q>0. This turns out to be true only for Z 82 isotopes (Pb and heavier). There are a few exceptions. Let s calculate the kinetic energy of an alpha particle in the decay P D+α. Start from the rest frame of the parent. From momentum conservation: m D v D = m v We can replace the velocity with the kinetic energy: r r 2KD 2K m D = m! m D K D = m K m D m This means that the disintegration energy is: Q = K D + K = m m D K + K = m + m D m D K 14
Alternatively we can write K = m D m + m D Q where Q =(m P m D m )c 2 where you can notice that this is an exact expression. Because mass is approximately proportional to the mass number A, we get K A 4 A Q where A is the mass number of the parent nucleus. Since A 4, K Q, which means that most of the kinetic energy is carried away by the α particle. 15
Alpha decay occurs because while the alpha particles are confined within the nuclear potential, occasionally they can sneak out. This occurs through a process known as quantum tunneling. Although classically, the alpha particle is not energetic enough to overcome the potential barrier, quantum mechanically, the wavefunction has some small probability to be found outside. 16
Beta decay: an electron and a neutrino are emitted: A ZP! A Z+1D+e + e Here, a neutron in the parent turned into a proton, and electron, and a neutrino, but only the electron and neutrino are emitted. The neutrino has no electric charge. It has (almost) zero mass. It (almost) never interacts with anything. How do we know it exists? 17
Suppose that we don t know about the neutrino. By analogy with our discussion of alpha decay, we have: Q =(m P m D m e )c 2 Beta decay will occur whenever mp>md. It is very common, and occurs for both light and heavy nuclei. What is the kinetic energy of the electron? K e = m D m e + m D Q Q (since m e m D ) If there is no neutrino, red is what is expected. Blue was what was observed. 18
Dear Radioactive Ladies and Gentlemen, As the bearer of these lines, to whom I graciously ask you to listen, will explain to you in more detail, how because of the "wrong" statistics of the N and Li6 nuclei and the continuous beta spectrum, I have hit upon a desperate remedy to save the "exchange theorem" of statistics and the law of conservation of energy. Namely, the possibility that there could exist in the nuclei electrically neutral particles, that I wish to call neutrons, which have spin 1/2 and obey the exclusion principle and which further differ from light quanta in that they do not travel with the velocity of light. The mass of the neutrons should be of the same order of magnitude as the electron mass and in any event not larger than 0.01 proton masses. The continuous beta spectrum would then become understandable by the assumption that in beta decay a neutron is emitted in addition to the electron such that the sum of the energies of the neutron and the electron is constant... I agree that my remedy could seem incredible because one should have seen these neutrons much earlier if they really exist. But only the one who dare can win and the difficult situation, due to the continuous structure of the beta spectrum, is lighted by a remark of my honoured predecessor, Mr Debye, who told me recently in Bruxelles: "Oh, It's well better not to think about this at all, like new taxes". From now on, every solution to the issue must be discussed. Thus, dear radioactive people, look and judge. Unfortunately, I cannot appear in Tubingen personally since I am indispensable here in Zurich because of a ball on the night of 6/7 December. With my best regards to you, and also to Mr Back. Your humble servant, W. Pauli 19
While alpha decay and gamma decay can be understood through existing physical phenomena (namely, quantum tunneling and excited nuclei, respectively), beta decay cannot. Beta decay requires a separate force for a complete explanation. This force is called the weak force (because it is weak). It s properties we will discuss in a few weeks time. 20