Radioactivity and energy levels Book page 497-503
Review of radioactivity β ; Free neutron proton β- decay is continuous β : Proton in nucleus neutron antineutrino neutrino
Summary of useful equations Rate of decay: dn = λn dt Activity: A = dn = λn or just A = λn dt A = A 0 2 n Decay equation: N = N 0 e ;λt N = N 0 1 2 n were n = t T 1 2 Half life: T1 2 = ln 2 λ Fraction remaining after time t: e ;λt This means e ;λt 1 2 = 1 cgrahamphysics.com 2 2016
Definitions Half life The time it takes the activity of a sample of the element to half in value Or The time it takes for half of the atoms in the sample of the element to decay Do NOT use mass Decay constant The probability of decay of a nucleus per unit time Radioactive decay Random and unpredictable The more radioactive nuclei are present the greater the probability of some decaying
Example Radium 226 emits alpha particles. The decay constant is 1.35 10 ;11. What mass of radium 226 is needed to give an activity of 2200Bq? Solution A = λn N = A = 2200 λ 1.35 10 11 = 1.63 1014 n = N N A = 1.63 1014 6.02 10 23 = 2.7 10;10 mol m = nm = 226 10 ;3 2.7 10 ;10 = 6.1 10 ;11 kg
Example A laboratory prepares a 10μg sample of caesium -134. The half life of caesium 134 is about 2.1 years. a) determine, in seconds, the decay constant for this isotope b) calculate the initial activity of the sample c) calculate the activity of the sample after 10.0 years Solution A) T1 2 = ln 2 λ λ = ln 2 T 1 2 B) n = m M = 10 10 6 134 = 0.693 2.1 365 24 3600 = 1.05 10;8 s ;1 = 7.5 10 ;8 mol N = nn A = 7.5 10 ;8 6.02 10 23 = 4.49 10 16 A = λn = 1.05 10 ;8 4.49 10 16 = 4.7 10 8 Bq C) A = A 0 e ;λt = 4.7 10 8 e ;0.33 10 = 1.7 10 7 Bq or 0.33 years
Measuring radioactive half life How to measure depends on the length of the half life Long half life Known mass of sample Measure activity using Geiger counter Use decay equation Find decay constant λ= T1 2 Example A sample of isotope of uranium -234 has a mass of 2.0 μg. Its activity is measured 3.0 10 3 Bq. What is its half life? Solution Given: A = dn dt = 3.0 103 Bq N = m N A = 2.0 10 6 6.02 10 23 = 3.3 1016 atoms dn dt N = 3.0 103 3.3 10 16 = 9.0 10;14 s ;1 dn dt = λn = ln 2 λ = ln 2 9.0 10 ;14 s ;1 = 7.6 1012 s = 2.4 10 5 years
Very long half life Geiger counter cannot be used for very long T1 such as uranium 2 238, which has a half life of just under 4.5 billion years The rate of decay cannot be measured A pure sample of nuclide in a known chemical form needs to be separated, its mass measured and then a count rate taken From the reading the activity can be calculated by multiplying the count rate by the ratio of area of spere of radius equal to position of G M tube window area of G M tube window Determine decay constant from mass of specimen using the same method as for long half life
Short half life order of hours Measure number of decays in short period (minutes) at different time intervals Plot A versa time and find T1 2 Plot log A versa time Yield a straight line Gradient = λ L o g A from the graph time Gradient = - λ
Very short half life order of seconds Use ionization properties Place sample in tube Apply electric field across tube Radiation from source will ionize air Ionization current is set up Decay of ionization current can be displayed on an oscilloscope
Nuclear energy levels Radioactive decay provides evidence for nucleus having energy levels Emission of α or β particles by radioactive parent nuclei often leaves daughter nucleus in excited state Daughter nucleus emits one or more γ ray photons as it reaches ground state
α- decay Decay route of Americum 241 to Neptunium - 237 Each nucleus emits an alpha particle having one of three possible energies (there are more, but for this example we have 3) Depending on emitted α particle Neptunium can be in ground state or in an excited state It will decay into ground state by emitting a single photon or, in two steps, two photons α particles have quantized energies γ particles have also quantized energies Hence nucleus must have energy levels
Mechanism of α decay leaving nucleus α particles form as clusters 2p + 2n inside the nucleus Nucleons are in random motion but energy less than energy needed to escape the nucleus Strong nuclear force provides potential energy barrier α particles need to overcome this (electrostatic force will then accelerate it away from the nucleus) According to classical mechanics the alpha particles do not have enough energy to leave
α decay and quantum mechanics Ψ of α particles not localized to nucleus Allows overlap with potential barrier provided by strong nuclear force There is a finite but very small probability of observing the alpha particle outside the nucleus Some particles will tunnel out of the nucleus
Experiments show With a higher potential barrier and a greater thickness to cross, a nucleus will have a larger life time This explains very long half lives of uranium and polonium When the wave function is at its maximum the probability of tunneling is greatest α particles with specific energies are most likely to be emitted
β decay is continuous It is continuous because neutrino accounts for any energy difference between maxima β decay and the sum of γ plus intermediate β particle energies Pauli suggested that a third particle was emitted in the decay to conserve spin and angular momentum as well as mass energy conservation and momentum
Fictinous example of β : decay β ; decay β : decay neutron proton + v proton neutron + v The neutrino accounts for the continuous β spectrum