KE A = PE MAX 1/2M v 2 = k q1 q2 /R

Transcription

1 CHAPTER 13 NUCLEAR STRUCTURE NUCLEAR FORCE The nucleus is help firmly together by the nuclear or strong force, We can estimate the nuclear force by observing that protons residing about 1fm = 10-15m apart are held in place by a binding potential of BE~ V = k e 2 / R = V = k e 2 / R =( 8.9 x 10 9 N-m 2 /C 2 ) (1.6 c C ) 2 / (10-15 m) = 2.3 x J = 2.3 x J / 1.6 x J/eV = 1.4 MeV per pair NUCLEAR SIZE From Alpha Rutherford determined that the nucleus was very small in comparisom to the atom. The alpha particles scattering from a Au foils indicated that they deflected from a point-like positive charge at the atom s center. KE A = PE MAX 1/2M v 2 = k q1 q2 /R The radial distribution R depended on the atomic number A. Since the mass of the nucleus M N A = r [4/3 p R 3 ] (r is the nuclear density) we can conclude that R = [(3/4 p) (1/r)M N ] 1/3 A 1/3 = 1.2 A 1/3 fm More detailed studies showed resulted in a model in which the nucleus can be described by a radius R and surface thickness a~1/2fm. r R DR NUCLEAR MODEL r(r) = ro /1+ exp[(r-r)/a] R = ~ 1.07 A 1/3 fm r o = N/fm 3 DR = 4.4 a a = = 0.55 fm (surface thickness) R(fm)

2 LINE OF NUCLEAR STABILITY Z = N The nuclear force requires a balance between the number of protons and neutrons to a large extent. Some of this is due to the large binding energy of the alpha particle block p-p-n-n. As the atomic number increases more neutrons are needed to keep protons apart, thus reducing the Coulomb repulsion. 120 N N= Z 0 90 Z MAGNETIC MOMENT and NUCLEAR SPIN The magnetic moment m of a sample is defined as the magnetization M per unit volume, m = M / V. The external magnetic field at the is B = m o c M, where c is the magnetic susceptibility of the material. S M N B S N Prot The proton and neutron have an inrinsic magnetic moment m which is related to a quantity which elementary particle process called spin S.= ±1/ 2 and g = 2 for e,p,and n s. m = g m N S The quantity m N = e h/2m = 5.05 x J/T is called the nuclear magneton.

3 The energy of a magnetic dipole is given by U = -m B In and external magnetic field B along the z-axis the magnetic moment m orients along the z-axis up or down. Is a second RF field is applied with energy DE = hf = 2m N B the magnetic moment will absorb this energy and begin to oscillate. The excitation frequency must be exactly given by hf = 2m N B. B B (RF) q m Eo Eo + m N B Eo - m N B Example- What frequency do the oscillating magnetic dipoles emit? If B = 4.4 T find the resonant frequency. and hf = (5 x J/T) (4.4T) = 22 x J f = 20 x J / x J-s = 33.2 MHz MRI E&M waves from the 33.2 MHz oscillations are detected by pick up coils. An x-y image can be formed The B(z) field is set with a constant field Bo + trim coil fied db. Thus the B(z) varies for each x-y slices of the patient s body. The RF frequency f is modulated to look at slices such that the resonance condition hf = 2m N B. is strictly satisfied. Only tissue in the 4.4T x-y slice is detected. B(z) = 3.7T 3.8T 3.9T 4T 4.1T 4.2T 4.3T 4.4T 4.5 T 4.6T 4.7T Main Field Bo B(z) = Bo + (db/dz) Trim Coil db

4 BINDING ENERGY Binding energy is defined as the energy necessary to separate all the nucleons in the atom from each other. Earlier we defined this as the mass excess. BE = Z M(H) + N M(n) M( A X Z ) HYDROGEN P DEUTERIUM P-N BE = 2.2 MeV BE/A =1.1 MeV/N TRITIUM P-N-N BE = 8.5 MeV BE/A = 2.7 MeV/N HELIUM P-P-N-N BE = 27.5 MeV BE/A = 6.9 MeV/N IRON BE/A = 7.5 MeV/N Notice in the decay A -> B + C EXCHANGE FORCE MODEL One model for the nuclear force is that the proton and neutron are exchanging a A field of p mesons. This is called theyukawa model. n 1.5 fm p p p Virtual Particles QM existing for a short time, DE Dt h/2 p Some problems arise just as in the Bohr Model! m How can conservation of energy by applied. After many exchanges the nucleons would gain enough energy to fly apart. Under the QM uncertainty principle energy conservation can be violates if it happens quickly enough! DE Dt ~ h/2 Dt ~ h/2m p c 2 d max = c Dt = hc/m p c 2 = 1.2 A 1/3 fm = 1.5 fm We can estimate the pi meson mass as M p c 2 = hc/1fm = MeV/ 1.5 fm = 131 MeV/c 2

5 SEMI-EMPIRICAL MASS FORMULA BE = c1 A + c2 A 2/3 - c3 Z(Z-1)/ A 1/3 - c4 (N-Z) 2 /A MeV/A 9 7 BINDING ENERGY vs A Fe 56 c1 = 15.7 MeV c2 = 17.8 MeV c3 = 0.71 MeV c4 = 23.6 MeV + Nuclear Volume - Surface Energy Term - Coulomb Repulsion Term - (N-Z) asymmetry term A INDEPENDENT PARTILCE MODEL SHELL MODEL In this model the nucleons sit in a square well whose depth and width depends N and Z of the nucleus. Proton levels are slightly elevated with respect to neutrons due to the Coulomb replulsion term. In solving this problem the energy levels of a single nucleon is written with the PE term corresponding to the mean potential due to all other nucleons- Mean Field Approach. The magic numbers correspond to filled shells just as in the atom s periodic table. neutrons protons Magic Numbers Z = N = N=Z=28 N=Z=20 N=Z=8 N=Z=2 Alpha Block. Blocks. Very Stable. Alpha particles p-p-n-n are particularly stable in the nuclei and often stick together, eg. alpha decay.

6 RADIOACTIVE DECAY In the creation of the Universe Atomic Nuclei were under constant bombardment from other particles as p, n, b, a, g s. We say the nuclei are activated or excited. Upon activation some radioactive nuclei decay promptly within milliseconds or can remain radioactive for millions of years. Excitation p, n, b, g, a s. Nuclei Prompt De-exitation p, n, b, g, a s t~1 s Slow De-exitation p, n, b, g, a s t ~[1s -1Myrs] MEAN LIFE - t and DECAY RATE l =1/t The radioactive decay of a nuclei follows an exponential decay law with mean lifetime t. Thenumber of surviving nuclei in a sample after a time t is given by N(t) = N(0) exp(- l t). The number decaying is then N D (t) = N(0) N(t) = N(0) [1 - exp(- l t) ] The decay rate R (t) = dn/dt = ( N(0) l ) exp(- l t )

7 HALF LIFE We can ask at what time will the decay rate drop by a factor of 2. Or when will the number of decaying nuclei drop by a factor of 2? This time t = T 1/2 (HALFLIFE) is given by N(0)/2 = N(0) exp(- l t ) or exp(- l t ) = 1/2 - l t = ln(1/2) = -ln(2) T 1/2 = 0.693/l = t N(t) = N(0) (1/2) t / T1/2 UNIT OF ACTIVITY The unit of radioactive decay is given by the number of decays per second coming from 1 gram of radium and called the curie (Ci): 1 Ci = 3.7 x decays/s A more convenient unit the Becquerel (Bq) is commonly used 1 Bq = 1 decay/s and 1 Ci = 3.7 x Bq. QUIZ on Thursday- Example 13.6!! 35 mg of pure 11 C 6 has a half life of 20.4 min. Determine the number of nuclei in the sample t =0. a, b, g DECAYS n p a e- g n a Decay A X Z A-4 Y Z a 2 b Decay A X Z A Y Z e n 0 g Decay A X Z * A X Z + 0 g 0

8 DECAY PROCESSES Y X a Alpha Decay An alpha particle tunnels through the nuclear barrier with decay rate l = R = 1/t The Parent X decays to Daughter Y plus an alpha particle. The available kinetic energy is A X Z A-4 Y Z a 2 Q = ( M X M Y -M a ) c 2 N a The alpha particle receives most of this Q in terms of kinetic energy. KE a = M Y / (M Y + M a ) Q KE a Beta Decay Beta decay is a 3-body decay process in which a neutron in the nucleus decays to a proton, electron, and anti-neutrino. (n p + e + n ). M n = 0!! A X Z A Y Z e n 0 Q = ( M X M Y M e ) c 2 <KE e > = Q / 3 KE MAX ~ Q 2-body alpha decay spectrum. Monoenergetic! N b 3-body decay spectrum of electrons. <KE e > ~ Q / 3 KE MAX = Q e Y e n Y n Neutrino at rest! Inverse beta decay p n + e + v (positron emission) Electron capture p + e - n + v

9 Gamma Decay An excited nucleus decays to a more stable form through emission of a gamma ray. Very similar to transititions in the hydrogen atom! A X Z * A X Z + 0 g 0 Eg = E* - E Excitation Processes Collisional Excitation A + B A + B* Post Alpha A -> B* + a Post Beta Decay A -> B* + e + n 12 B 5 12 C 6 * b- 12 C 6 g b- 15.4MeV 4.4MeV 0.0MeV CARBON DATING. Cosmic Rays create neutrons in the upper atmosphere for billions of years creating a constant fraction f(0) = N( 14 C)/N( 12 C) = 1.3 x The C forms in to molecules including CO 2 which plants breath. n + 14 N 7 -> 14 C 6 + p All living plants and animals acquire 14 C in their bodies. Upon death 14 C is not replenished and beta decays with a lifetme of t = 8270 yrs. 14 C 6 14 N b 1 + n T1/2 = 5730 yrs By measuring the ratio of 14 C/ 12 C in a present day we can estimate the age of the sample. f(t) = f(0) x e - t / T1/2

10 C14 atoms C14 atoms per gram RADIOCARBON DATING Example 13-11: YEARS A 25g of charcoal is found in an ancient city. 250 C14 decays per minute are measured. How old is the charcoal? l = 1/t = (1/8270) yrs =3.8 x s -1 N(C 12 ) = (25/12)mol N A = 1.26 x nuclei N O (C 14 ) = 1.3 x N(C 12 ) = 1.6 x nuclei R O (C 14 ) =l N O (C 14 ) = 6.13 decays/s = 370 decays/min (250/370) = e -l t t = 3200 yrs

11 NATURAL RADIOACTIVE SERIES The four naturally occuring radioactive series are: 238 U 92 -> -> -> 206 Pb 82 Uranium 235 U 92 -> -> -> 207 Pb 82 Actinium 232 Th 90 -> -> -> 208 Pb 82 Thorium 237 Np 93 -> -> -> 209 Bi 83 Neptunium These decay sequences occur through multiple a and b emissions down to Pb and Bi. ALPHA DECAY KINEMATICS A B + a B A a P B = P a = P E A = E B + E a M A c 2 = (KE B + M B c 2 ) + ( KE a + M a c 2 ) Q = (M A - M B -M a )c 2 = KE B + KE a Q = P B 2 / 2M B + P a 2 /2M a nonrelativistic approximation = 1/2 P 2 (M B +M a )/M B M a KE a = P 2 /2M a = [M B / (M B +M a )] Q

12