Chapter 31: Nuclear Physics & Radioactivity. The Nucleus

Transcription

1 Chapter 31: Nuclear Physics & Radioactivity Nuclear structure, nuclear size The strong nuclear force, nuclear stability, binding energy Radioactive decay, activity The neutrino Radioactive age measurement Decay series 1 The Nucleus Protons and neutrons ( nucleons ) are closely packed together in nuclei that are roughly spherical in shape. Proton:! q = +e Neutron: q = 0! neutrons and protons have almost the same mass Number of protons,! Z = atomic number Number of neutrons, N = neutron number Total number of nucleons, A = mass number, or nucleon number A = Z + N Nuclei are specified by: Example, 14 6C A ZX chemical symbol of the element Z is sometimes omitted, as the chemical symbol gives the same information

2 Isotopes: Nuclei of the same chemical element (same atomic number, Z), but different A and N. Example: 1 C, 13 C, 14 C. Only 1 C, 13 C are stable. The radius of a nucleus of mass number A is: r = r 0 A 1/3, r 0 = m Insofar as this is accurate, that means that nuclei have constant density: Density = mass volume Am N 4 3!r3 (m N = average mass of a nucleon) Density = Am N 4 3!r3 0 A = 3m N 4!r0 3 = kg 4!( m) 3 = kg/m 3!!! Comparable with the supposed density of a black hole or a neutron star. p n The Strong Nuclear Force Nuclei are held together by the strong nuclear force. gravity is much too weak the Coulomb force between proton! charges is repulsive and decreases! nuclear stability. Stable nuclei Effect of Coulomb repulsion between protons The strong nuclear force is: attractive felt over only ~10-15 m (a! short-range, nearest-neighbour force) The repulsive Coulomb force between protons favours nuclei with slightly more neutrons than protons. 4

3 Binding energy, mass defect Z protons + N neutrons Z protons, N neutrons Mass Zm p + Nm n!m = mass defect m nucleus Mass defect:!m = (Zm p + Nm n ) m nucleus Binding energy: B =!m c m nucleus = (Zm p + Nm n ) B/c Binding energy = energy to break up the nucleus into its constituent nucleons. Alternatively, a neutral atom with Z electrons is broken up into N neutrons and Z hydrogen atoms. Then,!m = [(Zm H + Nm n ) m atom ]. 5 Atomic and Nuclear Mass Atomic mass unit (u): 1 u = mass of 1 C atom (including the 6 electrons) 1 u is equivalent to a mass energy, mc, of MeV. Particle Electric Charge Kilograms Mass Atomic Mass Units (u) Electron -e " "10-4 Proton +e 1.676" Neutron " Hydrogen atom " Atomic mass (including Z electrons) = nuclear mass + mass of Z electrons. 6

4 Example The mass defect is:!m = " (mass of H atom + mass of neutron) (mass of 4 He atom)! = u!m = u Using the energy equivalent of the atomic mass unit, the binding energy is: B = ( u) " MeV/u = 8.4 MeV. Mass defect,!m = B c = J ( ) = kg 7 Binding energy per nucleon, B/A Fission " energy release Peak value, 8.7 MeV per nucleon Sharp fall due to small number of nearest neighbour nucleons short range nuclear force Unstable beyond 09 Bi Decrease due to Coulomb repulsion between protons Why are certain nuclei unstable? Because neighbouring nuclei have lower mass energy. Decay is possible to the lower mass nuclei while releasing kinetic energy

5 Radioactivity Three forms: Alpha (#)! the nucleus of a 4 He atom is emitted from the parent nucleus Beta ($)! an electron (+ or charge) is emitted Gamma (%) a nucleus falls from one energy level to another and emits a!! gamma ray photon 9 Nucleon number Charge Alpha Decay A ZX A 4 Z Y + 4 He Parent Daughter (#) Nucleon number: A = (A 4) + 4! Charge:! Z = (Z ) +! 38 9U 34 90Th + 4 He Daughter and #-particle: greater binding energy, lower combined mass than parent " energy is released in the decay. Energy released = [m X (m Y + m # )] " MeV, if masses in atomic mass units (u). The #-particles have a kinetic energy of typically a few MeV. 10

6 Alpha Decay in a Smoke Detector Alpha particles from a weak source collide with air molecules and ionize them, which allows a current to flow between the plates. In the presence of smoke, ions colliding with smoke are generally neutralized (i.e. neutral atoms are formed), so that the current decreases and the alarm is tripped. When the battery is low, the current is low, which also trips the alarm! Prob. 31.0/50: Find the energy (in MeV) released when alpha-decay converts radium 6 Ra (Z = 88, atomic mass = u) into radon Rn (Z = 86, atomic mass = u). The atomic mass of an alpha particle is u. 1

7 Beta ($ ) Decay Nucleon number Charge A ZX Z+1Y A + e Nucleon number: A = A + 0 (or $ ) Charge:! Z = (Z + 1) + ( 1) Nucleon number Charge Positron ($ + ) Decay A ZX 34 90Th 34 91Pa + e Z 1Y A + e + Nucleon number: A = A + 0 Charge:! Z = (Z 1) + 1 (or $ + ) 11Na 10Ne + e + 13 Beta-decay a problem Beta-decay X! Y + e Y (daughter) X (parent, at rest) Expected energy of the e + e or e + Kinematics: if energy and momentum are conserved, the electron (e or e + ) should have a well-determined kinetic energy following the beta-decay. But, the electron does not have a well-determined energy, as seen above. Is energy not conserved?! No, energy is conserved... 14

8 The Neutrino A third particle, a neutrino, is also emitted in the decay, so that the released energy is shared between three particles instead of two: 64 9Cu 64 8Ni + e + + ν Expected energy of the e + The neutrino is very difficult to detect. 15 Gamma (%) Decay A ZX A ZX +! Nuclear Energy X * X Excited state of the nucleus % ray photon Gamma rays are produced in the decay (de-excitation) of a nuclear state. This is similar to the production of a photon by an atom, except that the energy levels are associated with the nucleus itself, not with electrons in the atom. Gamma rays are generally of higher energy and are even more penetrating than x-rays. 16

9 Gamma Knife to zap a tumour 60 Co gamma ray source: 60 7Co 8Ni 60 + e +! Tumour 60 8Ni 8Ni 60 +! ( 1. MeV) Gamma rays from 60 Co sources are channeled through collimation channels in a metal helmet. Gamma rays are concentrated at the site of the tumour, to selectively destroy the malignant tissue Winnipeg Free Press, April 4,

10 N 0 Radioactive Decay Start with N0 unstable nuclei Observe how many survive to time t Half-life, : the time for half of the nuclei to decay. N 0 / After each succeeding half-life, half of the remaining unstable nuclei remain... N 0 /4 N 0 /8 19 After time, [ 1 After time, [ ][ 1 1 Radioactive Decay ] N 0 are left ] N 0 = [ ] 1 N 0 are left After time 3, [ ][ ] 1 1 N 0 = [ 1 ] 3 N 0 are left After time n, So, N(t) = N 0 [ 1 [ ] n 1 N 0 are left ] n n = t = number of half-lives 0

11 Radioactive Decay So, N(t) = N 0 [ 1 ] n n = t = number of half-lives This is the same as an exponential decay: N(t) = N 0 [ 1 ] n = N 0 e!t & = decay constant Take logs: nln =!t t ln =! t So,! = ln = Natural log, log e and, n =! = t = = number of half lives elapsed 1 mean life 1 Isotope Some Half Lives Half Life Decay Mode 14 Po ms #, % 89 Kr 3.16 min $, % Rn 3.83 days #, % 60 Co 5.71 y $, % 90 Sr 9.1 y $ 6 Ra 1600 y #, % 14 C 5730 y $ 38 U 4.47"10 9 y #, % 115 In 4.41"10 14 y!!! $

12 Radioactive Decay, Activity Activity, A = number of decays per second =!N/!t, N = no. of nuclei left Statistically, the rate of decay is proportional to the number of radioactive nuclei present. A =!N!t = "N(t) = "N 0e "t A = dn Calculus initial activity, A 0 = & N 0 dt = d dt N 0e!t =!N 0 e!t Units: Becquerel (Bq): 1 Bq = 1 decay per second! Curie (Ci, old unit): 1 Ci = 3.7 x Bq = activity of 1 g of pure! radium Short half life means large decay constant and large activity (# = 0.693/T1/). 3 Radioactive Decay The number of radioactive nuclei left after time t is: N(t) = N 0 [ 1 ] n n = t = number of half-lives or N(t) = N 0 e!t,! = decay constant! = Activity, A(t) = &N(t) 4

13 A 38U A Radioactive Decay Series Rn Radon, of basement fame 38 9U 34 90Th + 4 He 34 90Th 34 91Pa + e +! and so on......ending at 06 8Pb Other decay series: 35 U 07 Pb 3 Th 08 Pb Z All formed in a supernova explosion about 4.5 billion years ago Radon in the Basement Rn produced in the decay chain that starts with 38 U. Rn half life is only 3.83 days, but it is generated continually by the decay of longer-lived nuclei. Suppose 3"10 7 radon nuclei are trapped in a basement when the walls are sealed so no more can enter. How many are left after 31 days? [ ] n 1 N = N 0, n = t = = ] = N 0 = N = N 0 [ 1 The initial activity of the radon is A 0 =!N 0 = N = s = 6.8 Bq

14 Prob. 31.3/34: Strontium 90 Sr has a half-life of 9.1 years. It is chemically similar to calcium, enters the body through the food chain and collects in the bones. Consequently, 90 Sr is a particularly serious health hazard. How long will it take for 99.99% of the 90 Sr released in a nuclear reactor accident to disappear? 7 Radioactive Dating The best known is radiocarbon dating, based on the decay of 14 C. Carbon-based life forms take up carbon in food or as CO (plants, trees, in photosynthesis). One atom in 8.3"10 11 of carbon has a 14 C nucleus, the rest are 1 C, 13 C. 14 C has a half-life of 5730 years, 1 C, 13 C are stable. When the organism dies, the uptake of carbon ceases, and the amount of 14 C present decreases, halving every 5730 years. Measure how much 14 C is left " work out how long since organism died. 8

15 003 Final, Q9: The ratio of the abundance of 14 C to 1 C in a sample of dead wood is one quarter the ratio for wood in a living tree. If the half life of 14 C is 5730 years, which of the following expressions determines how many years ago the wood died? a) "5730 b) 4"5730 c) 0.75"5730 d) 0.50"5730 e) 0.5" Prob : The practical limit for radiocarbon dating is about 41,000 years. What fraction of the 14 C is left after this time? (half-life = 5730 years) The number of half-lives that have elapsed in 41,000 years is: n = 41,000/5,730 = 7.16 and so the fraction of 14 C left after 41,000 years is N/N 0 = (1/) 7.16 = so, only 0.7% of the 14 C is left. 30

16 Radioactive Dating The amount of 14 C left can be measured by: Counting the rate of decay (activity) of 14 C the accuracy of the! age measurement depends on the size of the sample and on for how! long you are willing to count. The more decays seen, the more! accurate the measurement. Counting the number of 14 C nuclei directly by vaporizing the! sample and counting the 14 C nuclei in a mass spectrometer. You are! no longer waiting for nuclei to decay and can get much higher! precision on the age Radioactive dating origin of the 14 C The 14 C comes from cosmic rays that interact with 14 N in the upper atmosphere: n N 14 6C + p Stable, the usual form of N Number of nucleons: = Charge:! = The 14 C combines with O to form 14 CO which mixes with normal CO in a stable proportion (1 in 8.3"10 11 ). 14 C decays back to 14 N: 14 6 C 14 7N + e +! 3

17 Radioactive Dating on Geological Timescales 38 U " 06 Pb in decay series and = 4.5 " 10 9 years for 38 U. Number of #-particles produced in the decay series is (38 06)/4 = 8. If the decays occur in rock, the 4 He can be trapped. Measure how much 4 He is present in the rock. Each U decay should generate a total of 8 #-particles. Then, the original number of 38 U nuclei in the rock when it was formed is: N 0 = N(t) + N! N(t) = number of U nuclei present now 8 Now Now [ ] n 1 Then, N(t) = N 0, n = t t Oldest rocks, t = 3.7"10 9 y; meteorites, moon rocks, t = 4.5"10 9 y. + other methods based on ratios of isotopes. 33 Nuclei Made up of nucleons (neutrons + protons), and held together by the! strong nuclear force, which is of short range. Radius r = r 0 A 1/3, r 0 = 1."10-15 m. Binding energy, B = energy to separate neutral atom into neutrons and! hydrogen atoms. B =!m c,!m = mass defect. Unstable nuclei decay to objects of lower total mass, converting the! difference in mass to energy: A ' #-decay: ZX A 4 Z Y + 4 He + energy A! $-decay: ZX Z+1Y A + e +! + energy A ZX Z 1Y A + e + +! + energy A! %-decay: Z X A ZX +! + energy 34

18 Radioactive Decay Half life, = time for half of the radioactive nuclei to decay. Number of nuclei left after time t: [ ] n 1 N(t) = N 0 = N 0 e!t n = number of half-lives = t! = decay constant = Activity = rate of decay, A =!N/!t A =! N(t)! 1 Bq = 1 decay/second 35