PHY138Y Physics for the Life Sciences.

Transcription

1 PHY138Y Nuclear & Radiation Section. Worked Examples : A.W. Key Page 1 of 15 PHY138Y Physics for the Life Sciences. Nuclear and Radiation Section. Worked Examples. I have generated these examples to show you how the theoretical concepts described in the Supplementary Notes translate into practical calculations. I have used the standard suggested method of solving problems in the hope that, seeing how effective it is, you will be encouraged to continue to use it. The problems are numbered by the section in the Supplementary Notes which they most closely exemplify. I may work some of them in class. I expect there may be a few errors, mainly, I hope, of the arithmetical sort. A reward of $5 will be given to the sender of the first to point out an arithmetical error or an error in the physics or logic of the solutions (grammatical, stylistic, typos, rounding errors etc. don t count!). To receive this award, the happy winner must be seated in the main, ground floor section of Convocation Hall at the time of the announcement. 1.4(i) The Molar Mass of Sodium 23 What is the mass of 1 mole of 11 Na? You may assume that the binding energy of each constituent nucleon is MeV, and that the binding energies of the electrons can be neglected. The proton mass, m p = u, the neutron mass, m n = u, and the electron mass, m e = u Model: 1 mole is that quantity of the element that contains Avogadro s number of atoms. So to find the mass of 23 one mole of 11 Na we need to multiply the mass of each atom by N A. The mass of the atom is the sum of the masses of the nucleons and electrons less their binding energy in the nucleus; we can neglect the binding energy of the electrons. From the definition of a mole, we d expect the answer to be very close to 23g. The mass of a 23 Na atom = (11m p +12m n +11m e ) (the nuclear binding energies of the 11 protons and 12 neutrons in the nucleus) (the atomic binding energy of the 11 electrons in the atom). The binding energy of the 11 electrons is typically tens of ev, and can be neglected. (11m p +12m n +11m e ) = ( ) u The binding energy of the 23 nucleons is 23 x MeV = MeV The nuclear binding energy of MeV corresponds to a mass of MeV/c 2, or / u = u (since 1u = MeV/c 2 ). 23 Thus the atomic mass of Na = ( ) u = u. (check Knight s table) Therefore the molar mass (the mass of 1 mole) of 11 Na is equal to Avogadro s number multiplied by the mass of each atom, or u x N A, or, converting to grams (1 u = {1/N A } g ), = g. Assess: This is close to, but not EXACTLY, 23 g, as expected.

2 PHY138Y Nuclear & Radiation Section. Worked Examples : A.W. Key Page 2 of (ii) Number of Atoms in a Given Mass A gold foil has a mass of 15 g. How many atoms are there in this foil? Model: We know that a mole of an element contains N A atoms. So we have to find out how many moles there are in 15g of Gold. Very closely, one mole of an element equals the mass number in g. The mass number of gold is 197. Less than N A ; about 1/197 of N A, so around Let m be the mass of the gold foil. m = 15 g. 1 mole of Au, containing N A atoms, has a mass of 197g. Thus there are N A /A atoms in one gram. So number of atoms in m g = m N A /A. No. of atoms in foil = mn A /A = ( atoms.g -1 ) (15 g) / 197 = atoms 1.5 The Hydrogen Spectrum In the Hydrogen spectrum, calculate the frequency and wavelength of A) the shortest visible wavelength line in the Balmer series, and B) the lowest frequency line in the the Paschen series. (Use the figure in SNI). A) the wavelength of the shortest visible wavelength line in the Balmer series Model: The shortest wavelength corresponds to the highest energy. The Balmer series corresponds to transitions to the n=2 level of Hydrogen. The figure in SNI, 1.5 shows that the maximum n in the Balmer series has n=6. The energy difference between the states with n=6 and n=2 is then equal to the energy of the emitted photon, hf. The visible spectrum has light of wavelength around 400 to 700 nm, so we d expect a few hundred nm. hf = (1/6 2 1/2 2 ) ev = 3.02 ev So the frequency, f = E / h = 3.02/ Hz = Hz ; And the wavelength λ = c / f = m = 411 nm (you can identify this line in the photograph of the Hydrogen spectrum shown in the figure) Assess: OK the Balmer spectrum is in the visible range (400 to700 nm). B): the wavelength of the lowest frequency line in the Paschen series Model: The lowest frequency corresponds to the lowest energy. The energy levels for the transition between levels with principal quantum numbers 4 and 3 can be read from the figure in SNI, 1.5. The Paschen series corresponds to transitions to the n=3 level of Hydrogen. From the figure it appears as if the energy differences in the Paschen series are less than the Balmer series. We might expect a lower energy, thus a longer wavelength. Thus hf = (-1.512) ev = ev So the frequency, f = E / h = / Hz = Hz : And the wavelength λ = c / f = m = 1890 nm = 1.89 µm. (this line doesn t appear in the photograph of the Hydrogen spectrum; why not?). Assess: OK

3 PHY138Y Nuclear & Radiation Section. Worked Examples : A.W. Key Page 3 of Bremsstrahlung X-ray. An X-ray tube operates at a voltage of 120 kv, and a tube current of 10mA. A) What is the power of the electron beam? B) Calculate the wavelength of the most energetic X-rays. A):The power of the electron beam? Model: Power is volts amps = watts or J. s -1. Note that the definition Power = current 2 resistance does not apply here, since the resistance of a vacuum tube is not a useful quantity for a vacuum tube. A light bulb has a power of about 100W so we d expect an X-ray tube to have considerably greater power. Guess 1000 W or more. P = VI, where V is the tube voltage, I is the current. So P = W = 1.2 kw. Assess :OK! B): the wavelength of the most energetic X-rays. Model: The most energetic X-rays are produced when all the electrons energy is given to the X-ray photon. The wavelength is inversely proportional to the frequency which, in turn, is proportional to the electron s energy. Thus a maximum energy corresponds to a maximum frequency which corresponds to a minimum wavelength. We also remember that an X-ray tube that produces diagnostic X-rays has voltages around 100V. We remember that diagnostic X-rays have energies up to around 100keV, and wavelengths in the range 10 to 100 pm. (1 pm = m. ). Let (E e ) max be the maximum energy of the electrons and λ min be the minimum wavelength. By definition, 1 ev is the energy acquired by an electron accelerated across 1 V. So the energy of these electrons is (E e ) max =120 kev, by definition. Thus the photons that receive all of the electron s energy (i.e. those of maximum energy) have an energy = hf max = (E e ) max. Thus the wavelength of these photons is given by λ min = c/f max = hc / (E e ) max. It is convenient to quote h in units of ev.s. Since 1 ev = J, h = J.s = / J.s./(J/eV) = ev.s λ min = hc / (E e ) max = ( ev.s) ( m.s -1 ) / ( ev) = m = 10.3 pm Assess: OK!

4 PHY138Y Nuclear & Radiation Section. Worked Examples : A.W. Key Page 4 of Dose in Air Given that it takes 33.7 ev to produce one ion pair in air, calculate the dose in air, D air, in Grays, produced by an X-ray exposure of X R. (This is the formula quoted in SNII). Model:. The dose is the energy deposited in 1 kg (definition). The product of the energy required to produce an ion pair and the number of ion pairs produced by 1 R will yield the total energy deposited by 1 R. We know that 1 R produces C of charge in 1 kg of dry air (definition). Each ion pair has a charge equal to the electronic charge, so we can calculate the number of ion pairs. Hard to guess. We might remember that 1 R delivers about 1 rad, or about 10-2 Gy. Let E(ip) = 33.7 ev: and Q = C : N(ip) = the number of ion pairs : e = electronic charge. N(ip) = Q/e Then energy deposited in 1 kg of dry air by X R is D air = E(ip) N(ip) = X E(ip) Q / e Energy deposited by X R in 1 kg of dry air = X (33.7 ev) ( )/( ) = X (33.7 ev) ( J.eV -1 ) ( )/( ) = X J. Thus D air = X J. kg -1 = X Gy. Thus, X R will give a dose in air of D air = X Gy. Assess: So OK! 2.5(i) Photon Fluence A monoenergetic beam of 25 MeV X-rays used in therapy has an energy fluence rate of 120 mj.m -2.s -1. If the beam has a cross-sectional area of 5 cm 2, what is the photon fluence rate? Model: the photon fluence rate is the number of photons per unit area per second. This number equals the energy delivered per unit area per second divided by the energy of each individual photon. who knows! 10 to the many powers! Let φ be the required photon fluence rate, let the energy of each photon be e γ and let the power of the beam be I. Then φ = I/ e γ Substitiute: φ = I/ e γ = (120 mj. m -2. s -1 )/(25 MeV) = ( / ) (J/eV)(m -2. s -1 ) = ( )/( ) m -2. s -1 = m -2. s -1. Assess: OK, not ridiculous!

5 PHY138Y Nuclear & Radiation Section. Worked Examples : A.W. Key Page 5 of (ii) Resolution of X-rays Mammograms are used to search for cancerous tumours in women s breasts. The breast is held steady between two parallel plates and the X-ray taken. Tumours have a density about 120% of healthy tissue. What is the minimum size of tumour located in 7 cm of breast tissue that can be detected by 40 kev X-rays? Assume that X-ray films usually allow contrasts of the order of 2 % (i.e. the ratio of the exposures at the X-ray film of a path including the tumour and a path that traverses only healthy tissue must be less than 98% to be detectable.) Representation: A diagram would be useful here. See opposite. Model: Since this question concerns resolution, we use the linear attenuation coefficient to measure the percentage of the original beam that survives its passage through tumour and tissue. Since tumours are denser than tissue, we d expect that they will attenuate the X-rays more than the surrounding tissue. The intensity (fluence) of a beam of X- rays decreases exponentially with depth. We have tables for healthy tissue in SNII, but are not given a value for the linear attenuation coefficient for the tumour. However, we know that the mass attenuation coefficient is approximately equal for a variety of tissues at X-ray energies. Obviously less than 7 cm! Optimistically we might hope for a few mm. Let s see! X Rays Breast Tissue Tumour FILM? 7cm Let s use photon fluence as our measure of the radiation reaching the X-ray film. Fluence decreases with distance as Φ(x) = Φ(0) exp( µx). Let d t be the total depth of healthy tissue : Let d c be the depth of the cancerous tumour Let µ t, ρ t be the linear attenuation coefficient and the density of healthy tissue respectively Let µ c, ρ c be the linear attenuation coefficient and the density of cancerous tissue respectively Let p% be the contrast limit: p = 98 The table in SNII lists the mass attenuation coefficients for tissue. The linear attenuation coefficients of healthy and cancerous tissue can then be found by multiplying by the density of each: µ t = (µ/ρ) t ρ t µ c = (µ/ρ) c ρ c Let s assume that, to a good approximation, the mass attenuation coefficient of healthy tissue equals that of cancerous tissue: (µ/ρ) t = (µ/ρ) c = cm 2.g -1 at 40 kev (see the table in SNII) The photon fluence emerging from a path that includes only healthy tissue is: Φ t (x) = Φ(0) exp( µ t d t ). The photon fluence emerging from a path that includes both healthy and cancerous tissue is: Φ t+c (x) = Φ(0) exp{ µ t (d t d c ) µ c d c } The ratio Φ t+c (x) / Φ t (x) must have a maximum value of p/100: Φ t+c (x) / Φ t (x) p / 100 = exp{ µ t (d t d c ) µ c d c + µ t d t } = exp{ µ t d c µ c d c } = exp (µ t µ c ) d c = exp{(µ/ρ) t d c (ρ t ρ c )}, setting (µ/ρ) t = (µ/ρ) c (µ/ρ) t d c (ρ c ρ t ) ln (100/p) d c (1.2 1) 0.95} ln (100/98) d c / ( ) 4.4 mm Assess: The result is not unreasonable though in practice, with several exposures, mammograms can reach resolutions of close to 0.1 mm.

6 PHY138Y Nuclear & Radiation Section. Worked Examples : A.W. Key Page 6 of (i) Radioactive Decay Most isotopes of calcium are stable. 49 Ca is an exception, with a half-life of 8.8 minutes. A) How many atoms of 49 Ca are there in a mass of calcium which has an activity of 260Bq? B) What is the mass of 49 Ca contained in this sample? A): No. of Atoms. Model: Use the formula R = λn, with decay constant λ = ln2 /T ½ = 0.693/ T ½ No. of atoms, N = R/λ = R T ½ / = ( /0.693) s -1.min. 60 s.min -1 = (NB the conversion to seconds). B): The mass of this sample of 49 Ca Model: Use Avogadro s number. Let M be the required mass. We know that N A atoms have a mass of (very closely) A= 49 g. So mass of N atoms = M = N A/N A = / g = g 3.2 (ii) Radioactive Decay 1 g of Radium ( 226 Ra) has an activity of 1 Curie (this was the definition of the Curie!). The half-life of 226 Ra is 1620 years. A) What will be the activity, in Bq, of 0.2g after 1620 years? B) What will be the activity of 0.2g after 2000 years? Model: The activity of a sample of a radioisotope is clearly proportional to the number of atoms, which is, in turn, proportional to the mass. Use R(t) = R(0) exp(-λt), with λ = ln2 /T ½ = 0.693/ T ½ We d expect then about 0.2/1 or 20% of the activity, or 20% of Bq, or about Bq to start. After one half-life, we would expect half of this value. Let M = 1g, m = 0.2 g, R(0) = 1 Ci = Bq, r(t) = activity of m at time t. Also decay constant λ = ln2 /T ½ = 0.693/ T ½ A): The activity after 1620 years? Define r(0) = mr(0)/m = ( / 1) Bq = Bq Then r(1620) = r(0)/2 = Bq Assess: OK! B): The activity after 2000 years? Even less but not a huge amount less. After 2000 years, r(2000) = r(0) exp(- ln2 2000/1620) = Bq = Bq Assess: OK!

7 PHY138Y Nuclear & Radiation Section. Worked Examples : A.W. Key Page 7 of Alpha Decay U emits an alpha, yielding Th. A) Calculate the disintegration energy of this decay. B) What is the kinetic energy of the emitted alpha? A): The disintegration energy. Energies in the nuclear realm are usually in the MeV range (binding energies per nucleon average around 8 MeV.) Model: The Q value is the energy available for the decay. We find it by calculating the difference between the mass of the parent nucleus and the sum of the masses of the decay nuclei. The tables give the masses of the atoms; the nuclear masses have to be calculated from the atomic masses by subtracting the masses of the electrons (the electronic binding energies can be neglected). Then convert this mass difference into energy using E = mc 2. Let M A be the atomic masses, and M N be the nuclear masses Q/c = M N ( U) - M N ( Th) - M N ( 2 He) = [M A ( U) 92 m e ] - [M A ( Th) 90 m e ] - [M A ( He) 2 m e ] = M A ( U) M A ( 90 Th) M A ( 2 He) [NB in this case the electronic masses cancel, so we can use the atomic masses directly] Do the arithmetic to obtain - Q/c 2 = u = ( ) u (MeV/c 2 )/u = 4.28 MeV/c 2 So Q= 4.28 MeV B): What is the kinetic energy of the emitted alpha? Only very slightly less than Q; since the alpha and the nucleus have equal but opposite momenta, p, and kinetic energy = p 2 /2m. Thus the massive nucleus carries off very little energy, leaving most for the alpha. Model: The momenta of the alpha and the daughter must be equal. We can express Q in terms of a ratio of energies. We then use energy = p 2 /2m, assuming that we do not need to use relativistic formulae. 4 Let M N ( 2 He) = m(α) 234 E( Th)/E(α) = [p /2 M N ( Th)]/ [p 2 /2m(α)] = m(α)/ M N ( 234 Th) Q = E(α) + E( 234 Th) = E (α) [ 1 + E( 234 Th)/E(α)] = E (α) [ 1 + m(α)/ M N ( So E (α) = Q/[ 1 + m(α)/ M N ( 234 Th)] Q/[ 1 + 4/234] 90 E (α) 4.28/[ 1 + 4/234] MeV = 4.20 MeV Assess: OK! Th)]

8 PHY138Y Nuclear & Radiation Section. Worked Examples : A.W. Key Page 8 of (i) Dose from X-rays of a given Exposure. The rule-of-thumb quoted in 2.4 claims that a whole-body exposure of 1R yields an absorbed dose of approximately 1 rad or 10 mgy. Verify! (This example gives a very approximate estimate of the dose from X-rays; the example that follows shows the details of a more careful calculation). Model: The dose in matter, being proportional to the energy fluence, decreases exponentially proportional to exp(-μ en x). For a rule-of-thumb the influence of the X-ray energy or the size or composition of the body being irradiated are clearly not important. Two assumptions spring to mind: 1. If we assume that all the energy of the X-rays is deposited in the body, the energy fluence at the exiting surface of the body has to be assumed to go to zero, which would require a body of infinite depth! 2. If we assume that the dose remains constant as the X-rays traverse the body, we ignore the exponential decay of the energy fluence. While this is not very realistic, it is simpler than previous assumption. So let s examine the problem using assumption 2. N.B. Note that in general, the dose in air is not equal to the dose in tissue, though the two are fairly close. In all of these calculations, we use the fact that all of the photons in the air strike the skin surface: this means that the fluence in air (number of photons per unit area) just outside the skin equals the fluence that reaches the skin. The exact formula for air is D air = X Gy (this comes directly from the definition of the Roentgen; see example 2.4 above ). Also, from 4.2.1, the energy fluence in air is given by Ψ = D air / (µ en /ρ) air. This energy fluence is equal to the energy fluence that enters the body, where Ψ = D m / (µ en /ρ) m (for the biological material m). Substitution yields D m = f m X, where f m = (µ en /ρ) m / (µ en /ρ) air. Using our assumption above, this is the required value of D m. f m has been calculated in the table from the data in the table in Since f m depends on energy for bodily materials, we would need to know the energy distribution of the X-rays, and also the composition of the body (how much tissue, muscle, bone, etc.) to make a detailed calculation. Such precision is clearly not required here. Note that only for bone at energies of less than 150 kev or so is f m significantly different from 10; since bone is approximately only 10% of body weight, we may take f m ~10 as an average for the whole body. Thus an exposure of 1R yields approximately 10 mgy (1 rad) about the same dose as is given to air. Air Tissue Muscle Bone Energy f m f m f m f m (kev) (mgy/r) (mgy/r) (mgy/r) (mgy/r) Assess: So the rule-of-thumb is confirmed.

9 PHY138Y Nuclear & Radiation Section. Worked Examples : A.W. Key Page 9 of (ii): Fluence, Dose, Mass Energy Absorption Coefficient. A beam of 20 kev X-ray photons delivers 1.3R to a 30 cm 2 area of a patient s skin. Calculate A) the energy fluence of the beam and B) the dose in air. C) at what depth of the patient s tissue does the energy fluence drop to below 5% of its entering value? D) What is the average dose delivered when this beam traverses this depth of tissue? A): the energy fluence of the beam Model: Assume that the X-ray beam is monoenergetic at 20keV (normally a poor assumption, but we are not given other instructions!). We know that the energy fluence is related to the exposure by Ψ = X / (µ en /ρ) air. Hard to guess here clearly if this is a realistic example, we d be worried if we got an answer of many Joules per m 2. So perhaps values of mj.m 2 or µj.m 2 would be OK. Ψ = X / (µ en /ρ) air, with X = 1.3 R from the table in SNIV, at 20 kev, for air, (µ en /ρ) air = cm 2.g -1 = (10-2 m) 2 /(10-3 )kg = m 2.kg -1. Thus Ψ = / = J/m 2 = 209 mj/m 2 (SI units in, SI units out!) Assess: On the high side, compared to our guess so check arithmetic particularly the conversions to SI units. B): the dose in air. Model: Dose is energy per kg. Use D air = X. We know that 1R gives a dose of about 10mGy in air (and in the body see problem 4.2.1(i). So we guess about 13mGy. Dose in air is D air = X with X = 1.3R. D air = = 11.3 mgy (again, SI units in, SI units out). Assess: Pretty much as expected. C): depth of tissue at which the energy fluence drops to below 5%? Model: Remember that the fluence in air just outside the skin is equal to the fluence that enters the body. The fluence decrease with depth as Ψ(x) = Ψ(0) exp( µ en x). The beam is of low energy, so will be expected to penetrate only a few cm. Let p be the percentage reduction p = 5 at a depth d. We want Ψ(d)/Ψ(0) = p/100, or exp( µ en d) = p/100. So µ en d = ln(100/p), and d = ln(100/p)/ µ en = ln(100/p)/ [ρ (µ en /ρ) tissue ] For tissue, (µ en /ρ) tissue = cm 2.g -1 at 20 kev, and ρ = So d = ln(100/5)/( ) = 9.7 cm Assess: OK! D): average dose delivered to the tissue up to this depth? Model: The dose is the energy deposited per kg. The energy deposited in a depth of tissue must equal the energy entering minus the energy leaving. The energy entering is found by multiplying the energy per m 2 (the energy fluence) by the area exposed. Then to find the average dose, divide the total energy deposited by the mass of the tissue exposed. less than 11.3 mgy (the dose in air) guess around 5 mgy. Let L be the depth (= 9.7 cm.), and let A be the area exposed (= 30 cm 2 ) Then Energy In = A Ψ(0) ; Energy Out = A Ψ(0) exp( µ en L) ; Energy deposited = A Ψ(0){1 - exp( µ en L)} If density of tissue is ρ, then mass of patient irradiated is ALρ So Average Dose = Dav = A Ψ(0){ 1 - exp( µ en L)}/ALρ = Ψ(0){ 1 - exp( µ en L)}/Lρ (N.B. A cancels). Density of tissue = 0.95 g.cm -3 = 0.95x10 3 kg.m -3 : Fluence Ψ(0) = 209 mj/m 2 : Depth, L =0.097 m : So Dav = (209 mj/m 2 ){1 exp(-0.325x0.95x9.7)}/ {(0.097 m)x (0.95x10 3 kg.m -3 )} = 209 x 0.95 / 92 mj.kg -1 = 2.15 mgy Assess: OK

10 PHY138Y Nuclear & Radiation Section. Worked Examples : A.W. Key Page 10 of , : (i) Equivalent dose, Effective Dose Suppose the whole body of a hospital worker is exposed to a 1R X-ray beam as shown in the first figure. A) what, approximately, is the Equivalent Dose and the Effective Dose? B) Suppose that shielding removes half of the X-ray beam, as shown in the second figure. What now is the approximate Equivalent Dose and Effective Dose? A): Equivalent Dose and the Effective Dose for full body exposure Model: Equivalent Dose is the Absorbed Dose multiplied by the Radiation Weighting Factor, W R, ( =1 for X-ray radiation). The Effective Dose is the Equivalent Dose multiplied by the sum of the Tissue Weighting Factors of the tissues irradiated. Since the whole body is irradiated, this sum, ΣW T = 1. Difficult to guess remembering that the average annual exposure from the environment is 3 msv, we d hope this poor person gets not more than several msv from what appears to be an accident. Equivalent Dose = 10 mgy 1 = 10mSv. Effective Dose = 10 1 msv = 10 msv B):Equivalent Dose and the Effective Dose for half body exposure Model: The Equivalent Dose measures the energy deposited per kilogram of biological tissue, so does not change for the half of the body exposed. However, the risk (of contracting cancer, etc.) has obviously decreased by a factor of two. Equivalent Dose = 10 msv, Effective Dose = 5 msv. Assess: OK! 4.2.2, : (ii) Equivalent dose, Effective Dose A patient has a 0.2 R chest X-ray. Estimate the Effective Dose. Model: We know that an exposure of 1R yields a dose of about 10mGy. W R =1 for X-rays. For a chest X-ray the X-rays penetrate breast tissue, muscle, bone. The bone traversed (ribs and spine) is a very small percentage of the total bone mass, which, in any case, has a value of W T of only Thus we choose to ignore the contribution from the bone. For the breast, W T = 0.15, and for the lung, W T = Equivalent Dose is msv = 2 msv. W T = = Thus the Effective Dose = msv = 0.54 msv. Assess: OK! Comment: To evaluate the possible risk to your health from this exposure, compare the exposure to your whole body from background radiation (about 3 msv in Toronto - see section 2.3.3). 4.3 Environmental Exposure Many foods contain potassium, a small fraction of which is a long-lived radioisotope. What is the radioactivity of your body from this source? (A 70 kg person has roughly 140 g of potassium, K, in the body. Most of this is the stable 39 K. However % of this mass is 40 K, the radioactive isotope with half-life of years; this emits beta-minus particles, of mean energy 561 kev.) Model: Knowing the mass of the potassium, we can calculate the number of radioactive atoms. The activity of the body is then given by R = λn, where λ can be calculated from the half-life. Such small energy beta rays can be assumed to be totally absorbed in the body. Must be small; most lab sources are around a few μci so several μci would be reasonable. Let T 1/2 be the half-life, λ the decay constant, and N = no. of atoms of 40 K. Let m be the mass of (stable) 39 K in the body, and let p be the small percentage of 40 K atoms in the 39 K. The mass of the 40 K atoms is (p/100) m, and the number of 40 K atoms in this mass = (p/100) m (N A /A) with A=40. Thus the activity of 40 K atoms = λn = (ln2/t 1/2 ) (p/100) (N A /A) m First calculate T 1/2 = yrs = s = s Thus Activity = (ln2/ [ s]) ( ) ([ /40] g -1 ) 140 g = 4231 Bq = µci Assess: OK! Comment. if you assume that % of the number of atoms of potassium are 40 K, the activity works out to be /39 = µci.)

11 PHY138Y Nuclear & Radiation Section. Worked Examples : A.W. Key Page 11 of Isotopic Dilution 100 μci of tritium, 3 H, was given to a person in the form of tritiated water that was totally absorbed through the gut. After time for equilibriation, a blood sample was taken that showed an activity of 92 Bq per ml of fluid. Calculate the total volume of water in the body. Model: the ratio of radioactive to stable isotopes is the same in the whole volume of water as it is in a small sample, assuming complete mixing. several litres? Initial activity, R(0) = 100 μci = Bq. Volume of sample = v = 1 ml = 10-3 l Activity of sample = r(0) = 92 Bq Then V = v R(0)/ r(0) = ( / 92) 10-3 l = 40 l. Assess: Perhaps this is a surprisingly large amount! In fact, the percentage of water in a baby s body is around 80%; it drops to around 60% for adults, so this problem yields a value that is, in fact, a bit low Committed Dose A 70 kg patient ingests 1.5µCi of 86 Rb that becomes uniformly distributed in the body. 86 Rb has half-life of 19 days and emits β and γ rays, with an average energy per disintegration of 0.60 MeV. Biological excretion is negligible. If 80% of this energy is retained in the patient s body, what is the committed dose he receives? Model: The committed dose is the total absorbed dose received over the entire lifetime of the radioisotope in the body. In this case this is just the total number of radioactive atoms deposited in the body multiplied by the energy each emits per disintegration, divided by the mass of the body. If this is a realistic example, we would hope that the total dose received by the patient is small compared to the normal annual dose of 2-3 msv per year. Perhaps a fraction of a mgy would be reasonable. Let M be the mass of the patient s body, N(0) the number of Rubididum atoms ingested, R(0) the activity, λ the decay constant, T 1/2 the half-life, and e n the energy of each disintegration. Let p be the percentage of the energy retained in the body. The number of atoms N(0) = R(0)/ λ = T 1/2 R(0)/ln2 Total energy released by these atoms = e n N(0) = e n T 1/2 R(0)/ln2 Total amount retained in body = (p/100) (e n T 1/2 R(0)/ln2) Committed Dose = (p/100) (e n T 1/2 R(0)/ln2)/M Committed Dose = 0.80 ( J) ( s) ( s -1 )/( kg) = J/kg = 0.14 mgy Assess: Looks OK.

12 PHY138Y Nuclear & Radiation Section. Worked Examples : A.W. Key Page 12 of Biological Excretion The half life for biological removal of phosphorous from the liver is 18 days. The nuclear half life of 32 P is 14.3 days. 32 P decays by beta-minus emission, with an average energy of 695 kev, all of which is absorbed in the liver. The liver has a mass of 1.7 kg. A) If there is 10 µci of 32 P in the liver at t=0 what will be the activity in the liver after 10 days? B) What is the dose delivered to the liver in these 10 days? A):What will be the activity in the liver after 10 days? Model: Radioactive 32 P decays with time due to the natural radioactive decay and normal excretion processes. We need to calculate the effective decay constant. The excretion speeds up the rate at which the radioisotope is removed, so we d expect the effective half-life to be less than 18 days and less than 14.3 days, say 10 days. So about half of the activity might exist in the body after these 10 days, or around 5 µci. 1/T eff = 1/T n + 1/T be ; so 1/T eff = 1/18 + 1/14.3 = 0.125, so T eff = 7.97 days Activity at time t = 10 days is given by R(10) = R(0)exp(-λ eff t) = R(0)exp( / T eff ) = 4.19 µci. Assess: Looks OK. B):What is the dose delivered to the liver in these 10 days? Model: Each nucleus that decays contributes 695 kev of energy. So we find the number of nuclei that have decayed in the body in 10 days, multiply by this energy per decay, and divide by the liver mass to get the absorbed dose. As usual, we might expect mgy or even μgy. N(0) = R(0)/ λ n and N(10) = R(10)/ λ n. Thus the number of nuclei that have either decayed or been excreted = N(0) - N(10) = [R(0) R(10)] / λ n. N.B. The ratio of those which decayed before they were excreted to those which were excreted is given by λ n / λ eff (see 5.4.2, where it is noted that dn n (t)/dn eff (t) = λ n /λ eff ). Thus the energy deposited by these nuclei = e β (λ n / λ eff ) [R(0) R(10)] / λ n = e β [R(0) R(10)] / λ eff where e β = 695 kev is the energy of each disintegration. Thus dose, D m (10) = e β [R(0) R(10)] T eff /(0.693 m), with m = 1.7 kg, the mass of the liver. D m (10) = ( ) [( ) ] ( )/( ) Gy D m (10) = 14.0 mgy. Assess: Looks OK.

13 PHY138Y Nuclear & Radiation Section. Worked Examples : A.W. Key Page 13 of Exposure from a Beam of Particles 15 mci of 99m Tc are drawn into a syringe. What is the exposure rate to a nurse standing 50 cm from the syringe, in mr.hr -1? The principal gamma ray from 99m Tc has MeV of energy, and the mass energy absorption coefficient of air for that energy is cm 2.g -1. Representation: A diagram may help here. Model: Treat the syringe as if it were a point source of gamma rays. The energy reaching a distance r goes as the inverse squared distance. There is also an exponential loss of energy as the gamma rays traverse the air. We d expect the former effect to dominate, but we should check that expectation. Effective Doses of less than 5 rem/year are considered probably safe or about 5 R/year, since 1 rad gives an exposure of about 1 R. 5 R/year 0.5 mr per hour [5/(365 24)]. So perhaps we d expect an answer around that value or somewhat higher after all the nurse isn t exposed to this all year! On the other 50 cm hand the exposure has to be large enough to provide enough radiation for the tracer to be useful. So a reasonable guess might be around 1 mr hr -1. The reduction in radiation at 50 cm is due to two effects: the absorption of energy [exp(-µ en x )] and the geometric effect [~1/distance 2 ]. Since the density of air is so low, the first factor can probably be ignored but let s check it! Let R(0) = activity, d = nurse to syringe distance, Mass energy absorption coefficient = (μ en / ρ ) air, e γ the energy of each gamma ray. Then the energy per second emitted from syringe = R(0) e γ J.s -1 and the energy per unit area = [R(0) e γ ]/ [4πd 2 ] J.s -1 m -2, since we assume that the gamma rays are emitted uniformly in all directions into a sphere of radius d. But this is just our old friend, the energy fluence rate, Ψ, which we know is related to the exposure X by the equation Ψ = X / (μ en / ρ ) air ( 2.2.1, 4.2.1) Activity = R(0) = 15 mci = s -1 Mass energy absorption coefficient: (μ en / ρ ) air = cm 2.g -1 = m 2.kg -1 Energy of photon = e γ =0.140 MeV = J = J Distance from syringe = d = 50 cm = 0.5 m (Now everything is in SI units). This must now be multiplied by exp(-µ en x ): but µ en = density of air; with the density of air = g.cm -3. so µ en = cm-1, and exp(-µ en x) = So, as we suspected, we can ignore the energy absorption in air. Thus, X = (μ en / ρ ) air Ψ / [ ] = [ (μ en / ρ ) air R(0) e γ ]/ [4π d 2 ]. (Notice that if we keep all units in SI, our answer should be in R.s -1 ). Inserting values, X = R.s -1 or mr.hr -1 = mr.hr -1. Assess:The guess is a bit low of course it is not being given for a whole year, so the answer is not way out of line). It is then but a short step to calculate the dose in air or in the person, using the other formulae in

14 PHY138Y Nuclear & Radiation Section. Worked Examples : A.W. Key Page 14 of Generation of Isototopes A Mo generator, of initial activity 1TBq is milked every 24 hours. What is the activity of the eluted Technetium 18 hours after the second elution? The half life of Mo is 2.5 days and the half-life of Tc is 6 hours. Representation: A diagram is useful here. Model: Immediately after each elution, the Tc activity has to start from zero again. So first calculate the activity of the Mo after two daily elutions, then adapt the formula R 2 (t) = R 1 (t) {λ 2 /( λ 2 - λ 1 )}{1 exp[-( λ 2 - λ 1 ) t]}. Remember to multiply by the branching ratio of from the diagram, after 2 days, 18 hours, the extrapolated graph gives an answer of around 40 mci for an initial Mo activity of 100 mci. So we might guess at an answer around 0.4 TBq. Activity (mci) Milking the Mo Cow (Transient Equilibrium) Time (hours) Let the parent Mo be milked after a time time T (= two days); its activity is then given by R 1 (T) = R 1 (0)exp(-λ 1 T). Then the Tc activity starting from zero at time T, begins to rise, with time t, according to R 2 (t) = R 1 (T+t) {λ 2 /( λ 2 - λ 1 )}{1 exp[-( λ 2 - λ 1 ) t]} = R(0)exp(-λ 1 [T+t]) {λ 2 /( λ 2 - λ 1 )}{1 exp[-( λ 2 - λ 1 ) t]}. In terms of half-lives, using λ 1 = ln2/t 1, etc., this expression becomes: R 2 (t) = R 1 (0)exp(-ln2 [T+t]/T 1 ) { T 1 /(T 1 - T 2 )}{1 exp[-ln2 ( T 1 T 2 ) /T 1 T 2 ] t}. T = 2 24 hrs : T 1 = hrs : T 2 = 6 hrs : t = 18 hrs : R(0) = 1 TBq = Bq : t + T = 66 hrs. R 2 (t) = exp( ) (0.115/0.104) (1 exp[ ]) TBq = TBq = 0.39 TBq Assess: OK! Mo Tc 6.2.3(i) Magnetic Resonance Imaging Explain in detail why MRI procedures are much safer for the patient than X-ray or Nuclear Medicine investigations. Model: The danger from X- and nuclear radiation arises from ionization in the tissues of the body. The energy required to ionize is typically greater than ev and kev or MeV radiation obviously can do even more damage. In MRI the only radiation that enters the body are the photons from the rf radiation. If their energy is small compared to the ionization energy, no damage will be done. Let s calculate! The typical (Larmor) frequency is given by f = γb 0 ~ 42B 0, with f in MHz and B 0 in T. Typical MRI magnetic fields are in the region of 1 to 5 T, so typical frequencies are 40 to 200 MHz. At the highest frequency, the energy of each photon is e γ = hf ~ 10-6 ev. Photons of this very low energy will do no damage to the human body! 6.2.3(ii) Magnetic Resonance Imaging In a very similar way to Nuclear Magnetic Resonance, spins of electrons in a material can be excited; the procedure is called Electron Spin Resonance (ESR). What is the equivalent Larmor frequency for electrons in a 1 T field? The magnetic moment of the electron is ev/t. Model: The Larmor frequency is given by υ L = γb 0 /2π, where γ is the gyromagnetic ratio; we need to calculate γ for the electron. The gyromagnetic ratio for the electron is given by γ = [μ /(½ ћ)]. So for the electron, γ = s -1 T -1. Thus υ L = /2π Mhz = Mhz. Comment: This is one reason why MRI uses protons rather than the (also plentiful) electrons. Human tissue is completely opaque to photons of this very high frequency.

15 PHY138Y Nuclear & Radiation Section. Worked Examples : A.W. Key Page 15 of (iii) Magnetic Resonance Imaging In an MRI procedure, the main field strength is B 0 = 1.5 T. If the applied rf field, B 1, has a strength of T (0.20 gauss). A) What should be the length of a 90 o rf pulse? B) How many oscillations of B 1 will have occurred in that time? A): What should be the length of a 90 o rf pulse? Model: The longer B 1 is applied, the greater the angle between the magnetization vector and the z-axis. Use the formula θ = γ B 1 τ. Short a few μs? τ = θ / (γ B 1 ) = (π/2)/( γ B 1 ) = (π/2)/( ) s = 290 μs Assess: longer than I d have guessed B): How many oscillations of B 1 will have occurred in that time? Model: each oscillation of the field takes one period and this period, T = 1/ υ L, where υ L is the Larmor frequency. The applied rf field frequency is of the order of the Larmor frequency, or 42 B 0 MHz = 63 MHz, with a period of T = 1/ υ, or 15.9 ns. In 290 μs, / = 18,300 oscillations of that field will have occurred in the time taken to reduce the z-component of the macroscopic magnetic moment of the protons to zero. 7 March 2008