UNIVERSITY OF CAPE TOWN. EEE4101F / EEE4103F: Basic Nuclear Physics Problem Set 04. Due 12:00 (!) Wednesday 8 April 2015

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1 UNIVERSITY OF CAPE TOWN EEE4101F / EEE4103F: Basic Nuclear Physics Problem Set 04 Due 12:00 (!) Wednesday 8 April ) Explain the concept of a cross-section, using language that a high-school teacher might understand, in less than a page of text. Do the reading first! You can do something like this: use the concept of the intensity of the beam (number of beam particles per unit area per unit time), the number of nuclei with which the beam particles can interact, and the rate (the number per second) of interactions that occur.... Also explain why the units of cross-section are those of an area.

2 2) A beam of 1.5 MeV neutrons, of intensity I 0 = cm 2 s 1 is incident on a slab of copper of thickness d = 0.75 cm. The emergent beam has intensity I = 0.23 I 0. Calculate the total cross-section of neutrons on copper at this energy. In a thick slab, the flux will have an exponential dependence on distance x. dφ = (µ dx)φ leads to Φ(x) = Φ(0) exp( µx). Consider a thick slab of thickness d. Let the front of the slab be at the origin x = 0. At the back of the slab x = d. Thus µ = ln(0.23)/(0.75 cm) =... = 1.96 cm 1 Let us use the definition of cross-section for a thin target (Lilley, Nuclear Physics p 24) in terms of the reaction rate R, the beam flux Φ and the number of target nuclei exposed to the beam N, R = N Φ σ = (I)(n t t)(σ) for a beam of area S, the beam intensity is I = Φ/S and a target of thickness t and number density of target nuclei n t = N/(St) If the target consists of nuclei of isotopic species with atomic mass M A (in atomic mass units), so M A = A, and the target density is ρ and N A is Avogadro s number, then n t = ρn A /A, and R = (I)(ρtN A /A)(σ) Now we have to relate σ to µ. Consider a thin slab of material. The incident flux is Φ in shines on an area S of the thin target which has thickness t. If the target is thin we can expand the exponential Φ out = Φ in e µt Φ in (1 µt) Φ in Φ out = Φ in (1 (1 µt)) = Φ in µt Now let us bring in σ. The reaction rate is R = NΦ in σ. Consider the rate per unit area R/S = (N/S)Φ in σ = (n t St/S)Φ in σ = (n t t)φ in σ This is just the flux that is removed from the beam. So (n t t)φ in σ = Φ in µt n t σ = µ Okay : σ = µ/n t. We look up the density of copper: ρ = 8.96 g cm 3. Copper has two abundant isotopes, so for A we take the atomic weight of natural copper, A = σ = µ/n t µ = ρn A /A (1.96)(63.5) = (8.96)( ) = cm 2

3 3) The total cross-section of thermal neutrons on carbon is 85 b. The elastic cross-section is by far the dominant component of the total cross-section. Calculate the mean free path of the neutrons. (See Lilley Ch 5 p143). As a beam of particles (say, here, neutrons) travels into a medium, its intensity decreases exponentially with distance I(x) = I 0 e λx. The probability of finding a neutron between x and x + dx is given by the weighted average of x, ie x exp( x/λ) dx 0 exp( x/λ) dx) = λ 0 The mean free path is the average distance the neutron travels before it has an interaction, and is thus equal to λ. Thus λ = 1/Σ where the macroscopic cross-section Σ = Nσ. Note that Lilley uses N to denote the number density of interacting objects. It is common to denote a number density by the use of the small letter n, and may authors do this. No if two sorts of interaction are possible, say scattering and absorption, then the total cross-section is given in terms of σ s and σ a by Σ = N(σ s + σ a = Σ s + Σ a and 1/λ = 1/λ s + 1/λ a Let us assume that the carbon is in the form of graphite. We look up the density of graphite and find ρ = 2.2 g cm 3. The number density of carbon nuclei ( 12 C is then n = N A ρ/a. Lilley uses N for n. λ = 1 Σ = 1 nσ = A N A ρσ (12) = =... = cm ( )(2.2)( )

4 4) The total mass attenuation coefficient for a 5.0 MeV photon in lead is cm 2 /g; in aluminum it is cm 2 /g. A source of such photons is shielded by a layer of lead and a layer of aluminum. The aluminum layer is 8 times the thickness of the lead layer. The photons pass first through the aluminum. What is the thickness of the lead layer if the photon intensity after passing through both layers is 1.5% of its initial intensity?

5 5) Use the ranges given in the figure below [Krane Figure 7.2] to compute the range, expressed in centimetres, of a) a 10 MeV α particle in gold; b) a 5 MeV proton in beryllium; c) a 1 MeV proton in water /...

6 Use the Bragg-Kleeman rule [An empirical relationship for estimating relative ranges of an ion in materials with different mass numbers (A) and densities (ρ)]: where R is the range in units of distance, or R 1 A1 ρ 2 R 2 ρ 1 A2 R 1 R 2 A1 A2 where R is the mass range in units of mass per unit area (like in Figure 7.2) (a) From the figure, the mass range for a 10-MeV α in Pb is about 39 mg cm 2. Using the R equation, A P b = u, and A Au = u, the mass range of a 10-MeV α particle in gold is mg cm 2, which divided by ρ Au = mg cm 3 gives a range of cm (b) From the figure, the mass range for a 5-MeV proton in Al is about 48 mg cm 2. Using the R equation, A Al = u, and A Be = u, the mass range of a 5-MeV proton in beryllium is mg cm 2, which divided by ρ Be = 1848 mg cm 3 gives a range of cm (c) For this one, it is not clear what material to use for comparison, so let s look at two option, H 2 and Al. - From the figure, the mass range for a 1-MeV proton in H 2 is about 0.8 mg cm 2. Using the R equation, A H2 = 2 u, and A H2 O = 18 u, the mass range of a 1-MeV proton in water is 2.40 mg cm 2, which divided by ρ H2 O = 1000 mg cm 3 gives a range of cm - From the figure, the mass range for a 1-MeV proton in Al is about 3.5 mg cm 2. Using the R equation, A Al = u, and A H2 O = 18 u, the mass range of a 1-MeV proton in water is mg cm 2, which divided by ρ H2 O = 1000 mg cm 3 a range of gives cm - The two materials produce slightly different values, since the actual value is cm, surprisingly H 2 produces a closer answer, and we can see that the method works reasonably well across different phases of matter.

7 6) The specific rate of energy loss ( de/ρdx) of a 5-MeV proton in silicon is 59 kev mg 1 cm 2 and its range R is 50 mg cm 2. Calculate values of ( de/ρdx) and range R for deuterons, tritons, 3 He and α particles, all of which have the same speed as the proton. For particles in the same material, the stopping power (from Bethe-Bloch) and range equations reduce to: de dx = z2 f(v) and R m z 2 F (v) depending only on the velocity v, the particle charge z, and the particle mass m. Since all the particles have the same velocity, the stopping power and range are related to our original proton simply by z and m, with final values listed the following table Particle z m de ρdx mg 1 cm 2 ) R (mg cm 2 ) proton ( 1 H) deuteron ( 2 H) triton ( 3 H) He alpha ( 4 He)

8 7) Calculate the thickness of cadmium (A = 112.4, density = 8650 kg/m 3 ) that would attenuate the intensity of a beam of thermal neutrons by a factor of The absorption cross section is 3000 b (1 b = m 2 ). The scattering cross section is negligible.

9 8) Calculate the average number of collisions needed for neutrons of initial energy 2.0 MeV to be thermalized (to average energy ev) in beryllium (A = 9) and iron (A = 56) and lead (A = 207). In each case, how many collisions would be needed to reduce the average energy to be about half the initial energy? (a) Using the atomic number for iron (A = 56), the logarithmic energy decrement, ξ, can be calculated: ( ) (A 1)2 A 1 ξ = 1 + 2A ln = A + 1 which can be used to calculate the number of collisions required to thermalize neutrons using (where E 0 = 2 MeV and E n = ev): n = 1 ξ ln E 0 E n = 516 collisions (b) Using the value of ξ = from above with E 0 = 2 MeV and E n = 1 MeV, the number of collisions is: n = 1 ξ ln E 0 E n = 20 collisions Full Results Material ξ n (to thermalize) n (to halve energy) Beryllium (A = 9) Iron (A = 56) Lead (A = 207) dga/swp/ solutions/dga/swp/