· Web view2021. 3. 14. · 3.8.1.4 Nuclear Instability. Q1. A rod made from uranium−238 (U)...

Colonel Frank Seely School

3.8.1.4 Nuclear Instability

Q1.A rod made from uranium−238 ( U) is placed in the core of a nuclear reactor where it absorbs free neutrons.When a nucleus of uranium−238 absorbs a neutron it becomes unstable and decays to

neptunium−239 ( Np), which in turn decays to plutonium−239 ( Pu).

(a) Write down the nuclear equation that represents the decay of neptunium−239 into plutonium−239.

(2)

(b) A sample of the rod is removed from the core and its radiation is monitored from time t = 0 s.The variation of the activity with time is shown in the graph.


(i) Show that the decay constant of the sample is about 3.4 × 10–6 s–1.

(2)

(ii) Assume that the activity shown in the graph comes only from the decay of neptunium.

Estimate the number of neptunium nuclei present in the sample at time t = 5.0 × 105 s.

number of nuclei .....................................................(1)

(c) (i) A chain reaction is maintained in the core of a thermal nuclear reactor that is operating normally.

Explain what is meant by a chain reaction, naming the materials and particles involved.

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(ii) Explain the purpose of a moderator in a thermal nuclear reactor.

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(iii) Substantial shielding around the core protects nearby workers from the most hazardous radiations. Radiation from the core includes α and β particles, γ rays, X−rays, neutrons and neutrinos.

Explain why the shielding becomes radioactive.

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(Total 11 marks)

Q2.(a) State what is meant by the binding energy of a nucleus.

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(b) (i) When a nucleus absorbs a slow-moving neutron and undergoes fission

one possible pair of fission fragments is technetium and indium .


Complete the following equation to represent this fission process.

(1)

(ii) Calculate the energy released, in MeV, when a single nucleus undergoes fission in this way.

binding energy per nucleon of = 7.59 MeV



energy released ..................................... MeV(3)

(iii) Calculate the loss of mass when a nucleus undergoes fission in this way.

loss of mass ......................................... kg(2)

(c) (i) On the figure below sketch a graph of neutron number, N, against proton number, Z, for stable nuclei.


proton number, Z

(1)

(ii) With reference to the figure, explain why fission fragments are unstable and explain what type of radiation they are likely to emit initially.

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(Total 12 marks)


Q3. The isotope of uranium, , decays into a stable isotope of lead, , by means of a series of α and β– decays.

(a) In this series of decays, α decay occurs 8 times and β– decay occurs n times.Calculate n.

answer = ...........................................(1)

(b) (i) Explain what is meant by the binding energy of a nucleus.

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(ii) Figure 1 shows the binding energy per nucleon for some stable nuclides.

Figure 1


Use Figure 1 to estimate the binding energy, in MeV, of the nucleus.

answer = ................................. MeV(1)

(c) The half-life of is 4.5 × 109 years, which is much larger than all the other half-lives of the decays in the series.

A rock sample when formed originally contained 3.0 × 1022 atoms of and no

atoms.

At any given time most of the atoms are either or with a negligible number of atoms in other forms in the decay series.

(i) Sketch on Figure 2 graphs to show how the number of atoms and the

number of atoms in the rock sample vary over a period of 1.0 × 1010 years from its formation.Label your graphs U and Pb.

Figure 2

(2)


(ii) A certain time, t, after its formation the sample contained twice as many

atoms as atoms.

Show that the number of atoms in the rock sample at time t was 2.0 × 1022.

(1)

(ii) Calculate t in years.

answer = ................................. years(3)

(Total 10 marks)

Q4. (a) Sketch, using the axes provided, a graph of neutron number, N, against proton number, Z, for stable nuclei over the range Z = 0 to Z = 80. Show suitable numerical values on the N axis.


(2)

(b) On the graph indicate, for each of the following, a possible position of a nuclide that may decay by

(i) α emission, labelling the position with W,

(ii) β– emission, labelling the position with X,

(iii) β+ emission, labelling the position with Y.(3)

(c) The isotope decays sequentially by emitting α particles and β– particles,

eventually forming the isotope . Four α particles are emitted in the sequence.

Calculate the number of β– particles in the sequence.


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(d) A particular nuclide is described as proton-rich. Discuss two ways in which the nuclide may decay. You may be awarded marks for the quality of written communication in your answer.

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(Total 10 marks)

Q5.(a) The diagram is copied from a photograph taken of a cloud chamber containing a small radioactive source.


(i) What type of radiation is emitted from the source?

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(ii) State and explain what can be deduced about the energy of the particles emitted by the source.

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(4)

(b) Plutonium-239 is a radioactive isotope that emits α particles of energy 5.1 MeV and decays to form a radioactive isotope of uranium. This isotope of uranium emits α particles of energy 4.5 MeV to form an isotope of thorium which is also radioactive.

(i) Write down an equation to represent the decay of plutonium-239.

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(ii) Write down an equation to represent the decay of the uranium isotope.

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(iii) Which of the two radioactive isotopes, plutonium-239 or the uranium isotope, has the longer half-life? Give a reason for your answer.


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(iv) Explain why thorium is likely to be a β– emitter.

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(Total 9 marks)

Q6.(a) (i) Sketch a graph to show how the neutron number, N, varies with the proton number, Z, for naturally occurring stable nuclei over the range Z = 0 to Z = 90. Show values of N and Z on the axes of your graph and draw the N = Z line.

(ii) On your graph mark points, one for each, to indicate the position of an


unstable nuclide which would be likely to be

an α emitter, labelling it A,

a β– emitter, labelling it B.(5)

(b) State the changes in N and Z which are produced in the emission of

(i) an α particle,

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(ii) a β– particle.

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(c) The results of electron scattering experiments using different target elements show that

where A is the nucleon number and r0 is a constant.

Use this equation to show that the density of a nucleus is independent of its mass.

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(Total 10 marks)


Q7. (a) (i) Explain why, despite the electrostatic repulsion between protons, the nuclei of most atoms of low nucleon number are stable.

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(ii) Suggest why stable nuclei of higher nucleon number have greater numbers of neutrons than protons.

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(iii) All nuclei have approximately the same density. State and explain what this suggests about the nature of the strong nuclear force.

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(b) (i) Compare the electrostatic repulsion and the gravitational attraction between a pair of protons the centres of which are separated by 1.2 ×10–15 m.


proton charge = 1.6 × 10–19 Cproton mass = 1.7 × 10–27 kggravitational constant = 6.7 × 10–11 N m2 kg–2

permittivity of free space = 8.9 × 10–12 F m–1

(ii) Comment on the relative roles of gravitational attraction and electrostatic repulsion in nuclear structure.

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(Total 11 marks)


M1.(a) Np → Pu + β - + ✓✓ First mark for one anti-neutrino or one beta minus particle in any form e.g. e-. If subscript and superscripts are given for these they must be correct but ignore the type of neutrino if indicated.The second mark is for both particles and the rest of the equation.Ignore the full sequence if it is shown but the Np to Pu must be given separately for the mark.

2

(b) (i) T1/2 2.0 2.1 × 105 s ✓then substitute and calculateλ = ln 2 / T1/2 ✓

T1 / 2 may be determined from graph not starting at zero time.Look for the correct power of 10 in the half-life ‒ possible AE.

Or(substitute two points from the graph into A = Aoe-λt)e.g. 0.77 × 1012 = 4.25 × 1012 exp(-λ×5×105) ✓then make λ the subject and calculate ✓(the rearrangement looks likeλ = [ln (Ao / A)] / tor λ = ‒ [ln (A / Ao )] / t )

Allow the rare alternative of using the time constant of the decayA = Ao exp (-t / ttc)from graph ttc = 2.9 3.1 × 105 s✓λ = 1 / ttc = 3.4 × 10-6 s-1 ✓No CE is allowed within this question.

both alternatives giveλ = 3.3 3.5 × 10-6 s-1 ✓

For referenceT1/2 = 2.0 × 105 s givesλ = 3.5 × 10-6 s-1 andT1/2 = 2.1 × 105 s givesλ = 3.3 × 10-6 s-1.

2

(ii) (using A = NλN = 0.77 × 1012 / 3.4 × 10-6 = 2.2(6) × 1017 )allow 2.2 2.4 × 1017 nuclei ✓


A possible route is find No = Ao / λthen use N = Noe-λt.Condone lone answer.

1

(c) (i) uranium (‒ 235 captures) a neutron (and splits into 2 smaller nuclei / fission fragments) releasing more neutrons ✓

First mark for uranium + neutron gives more neutrons.Ignore which isotope of uranium is used.

(at least one of) these neutrons go on to cause further / more splitting / fissioning (of uranium‒ 235) ✓

Second mark for released neutron causes more fission.The word ‘reactionߣ may replace ‘fissionߣ here provided ‘fission / splitting of uraniumߣ is given somewhere in the answer.

2

(ii) Escalate if clip shows critical mass in the question.the moderator slows down / reduces the kinetic energy of neutrons ✓so neutrons are absorbed / react / fission (efficiently) by the uranium / fuel ✓

owttePossible escalation.

2

(iii) neutrons are absorbed / collide with (by the nuclei in the shielding) ✓Second mark is only given if neutrons appear somewhere in the answer.

converting the nuclei / atoms (of the shielding) into unstable isotopes (owtte)

No neutrons = no marks.Making it neutron rich implies making them unstable.

2[11]

M2.(a) the amount of energy required to separate a nucleus ✓ into its separate neutrons and protons / nucleons ✓ (or energy released on formation of a nucleus ✓ from its separate neutrons and protons / constituents ✓)

1st mark is for correct energy flow direction


2nd mark is for binding or separating nucleons (nucleus is in the question but a reference to an atom will lose the mark)ignore discussion of SNF etcboth marks are independent

2

(b) (i) ✓must see subscript and superscripts

1

(ii) binding energy of U = 235 × 7.59 ✓ ( = 1784 (MeV)) binding energy of Tc and In = 112 × 8.36 + 122 × 8.51 ✓ ( = 1975 (MeV)) energy released ( = 1975 – 1784) = 191 (MeV) ✓ (allow 190 MeV)

1st mark is for 235 × 7.59 seen anywhere 2nd mark for 112 × 8.36 + 122 × 8.51 or 1975 is only given if there are no other terms or conversions added to the equation (ignore which way round the subtraction is positioned) correct final answer can score 3 marks

3

(iii) energy released = 191 × 1.60 × 10−13 ✓ ( = 3.06 × 10−11 J) loss of mass ( = E / c2 ) = 2.91 × 10−11 / (3.00 × 108)2) = 3.4 × 10−28 (kg) ✓ or = 191 / 931.5 u ✓ ( = 0.205 u) = 0.205 × 1.66 × 10−27 (kg) = 3.4 × 10−28 (kg) ✓

allow CE from (ii) working must be shown for a CE otherwise full marks can be given for correct answer onlynote for CE answer = (ii) × 1.78 × 10−30

(2.01 × 10−27 is a common answer)2

(c) (i) line or band from origin, starting at 45° up to Z approximately = 20 reading Z = 80, N = 110→130 ✓


initial gradient should be about 1 (ie Z = 20 ; N = 15 → 25) and overall must show some concave curvature. (Ignore slight waviness in the line) if band is shown take middle as the line if line stops at N > 70 extrapolate line to N = 80 for marking

1

(ii) fission fragments are (likely) to be above / to the left of the line of stability ✓ fission fragments are (likely) to have a larger N / Z ratio than stable nuclei or fission fragments are neutron rich owtte ✓ and become neutron or β− emitters ✓

ignore any reference to α emission a candidate must make a choice for the first two marks stating that there are more neutrons than protons is not enough for a mark 1st mark reference to graph 2nd mark – high N / Z ratio or neutron rich 3rd mark beta minus note not just beta

3[12]

M3. (a)

β = 6 1

(b) (i) the energy required to split up the nucleus

into its individual neutrons and protons/nucleons

(or the energy released to form/hold the nucleus

from its individual neutrons and protons/nucleons )2

(ii) 7.88 × 206 = 1620 MeV (allow 1600-1640 MeV)1


(c) (i) U, a graph starting at 3 × 1022 showing exponential fall passing through

0.75 × 1022 near 9 × 109 years

Pb, inverted graph of the above so that the graphs cross at 1.5 × 1022 near4.5 × 109 years

2

(ii) (u represents the number of uranium atoms then)

u = 6 × 1022 – 2u

u = 2 × 1022 atoms1

(iii) (use of N = No e-λt)

2 × 1022 = 3 × 1022 × e-λt

t = ln 1.5 / λ

(use of λ = ln 2 / t1/2)

λ = ln 2 / 4.5 × 109 = 1.54 × 10-10

t = 2.6 × 109 years (or 2.7 × 109 years)3

[10]

M4. (a) graph passes through N = 10/11 when Z = 10 and N increases as Zincreases (1)N = 115 → 125 when Z = 80 and graph must bend upwards (1)

2

(b) (i) W at Z > 60 just (within one diagonal of a square) below line (1)

(ii) X just (within one diagonal of a square) above line (1)

(iii) Y just (within one diagonal of a square) below line (1)3


(c) working showing the change due to emission of four α particles (1)four β– particles (1)

1

(d) Any two from the following list of processes:

β+

describe the changes to N (up by 1) and Z (down by 1)[or allow p change to n]

αmove closer to line of stability[or state the proton to neutron ratio is reduced]

ponly if nuclide is very proton rich[or electrostatic repulsion has to overcome the strong nuclear force][or highly unstable][or rare process]

e– capturedescribe the changes to N (up by 1) and Z (down by 1)allow p changes to n

marking: listing two processes (1)discussing each of the two processes (1) (1)

3QWC 1

[10]

M5.(a) (i) alpha (1)

(ii) two different track lengths (1)

short range particles have lower energy than long range particles (1)

particles in each range have same energy (1)(4)


(b) (i)

(ii)

(iii) U-235 (1)

because of the inverse relationship

between half-life and alpha particle energy (1)

(iv) because the Th-90 nucleus is neutron-rich compared

with U-235 [or Pu-239] (1)(5)

[9]

M6.(a)

(5)


(b) (i) α emitter: N ↓ 2, Z ↓ 2 (1)

(ii) β− emitter: N ↓ 1, Z ↑ 1 (1)(2)

(c) density = (1)

mass ∝ A

volume ∝ R3 and R ∝ hence volume ∝ A (1)

hence density = is independent of A (1)(3)

[10]

M7. (a) (i) strong nuclear force acts on all nucleons/both forces act onprotons/mention of gluons as force carrier

B1

strong nuclear force > electrostatic repulsion

B1

(ii) neutrons spread the protons out/neutrons reduceelectrostatic repulsion

B1

(iii) strong nuclear force has short range

M1

if snf fell off more gradually bigger nuclei would havelower densities/…more rapidly still higher densities

A1

strong nuclear force acts on all nucleons

M1


attractive nature of snf means all nucleons incontact/close packed

A1

strong nuclear force becomes repulsive at very smallseparations

M1

prevents nuclei from becoming denser

A1

needs minimum of two M1s to score all three heremax 6

(b) (i) or with k defined

C1

1.59 × 102N

A1

C1

1.3 × 10−34N

A1

(ii) can ignore gravitation when considering nuclearforces orgravitational force is much weaker than electrostaticforce not e.c.f.

B15

[11]


E1.This question was quite discriminating overall because of its synoptic nature and other mixed components. A majority of students got part (a) correct without too much difficulty. Those that did not, either missed off the antineutrino or they thought this stage of the decay was initiated by a neutron. Most students could perform the calculations required in part (b)(i). Most found the half-life and progressed from there. A significant number of successful students substituted data from the graph into a decay equation. Most of the students who succeeded in (b)(i) also succeeded in (b)(ii). The most common mistake was to leave out the power of 10 from the activity reading from the graph. Part (c)(i) caused students a number of problems. Many spent too much time saying what a chain reaction was in very general terms without reference to the specific situation. Many scripts started, ‘A chain reaction is when a process does something that creates an item that is needed for another process to take place...’ Usually this was given in a much more verbose fashion. When it did come down to the specifics students were not very careful about using the correct terms. Although not penalised here a majority who mentioned uranium used the 238 rather than the 235 isotope. It was also common to see words like react or decay being used where fission should have been used. Also when a single stage of the process had been written down the next stage was not explained in sufficient detail. The words, ‘and so on’ came far too early. In (c)(i) it was only the stronger students who knew the part played by the critical mass. These students tended to gain both marks available for this part question because they knew how it had an effect on the chain reaction. The majority of the other students thought the mass had something to do with the mass of individual nuclei and its effect on an individual fission process. The final part (c)(iii) was done poorly by all but the most able students. Most thought that the ionisation caused by radiation made atoms radioactively unstable. Very few were aware of the problems caused by exposure to a flux of neutrons.

E2.In part (a) was very straightforward for many candidates. Less able candidates did not elaborate on what the nucleus split up into or they referred to some other splitting such as fission. Another common error was to talk about the separation of an atom into protons, neutrons and electrons. Very few candidates got the direction of energy flow wrong.

In part (b)(i) very few candidates referred to particles other than neutrons but a significant number failed to write down two neutrons in the correct manner. and a single were extremely common.

The follow on part (b)(ii) turned out to be an extremely good discriminator. It highlighted the types of errors candidates were making. There was a group who did not appreciate the data was given per nucleon and used the figures without multiplying up by the respective nucleon numbers. Another significant group added the mass energy of mostly one but sometimes two nucleons into the proceedings. The last major group got into difficulties because they changed units unnecessarily or only changed the units of some of the terms. Overall there was a good percentage of correct answers.

A good proportion of candidates could accomplish the conversion of units required in part (b)(iii) with relative ease, sometimes from an error carried forward from the previous part. With the conversion needing two stages using the datasheet there was plenty of


opportunity for errors by dividing instead of multiplying or using an incorrect conversion factor. A separate but common error was to ignore the answer to part (b)(ii) and simply use the difference in mass of the nucleons involved, ignoring the binding energy per nucleon completely. Less able candidates did not really help themselves in this question because very few put words or units to intermediate stages of the calculation. If they had done so fewer would have lost their way.

A majority of candidates did not score marks in part (c)(i). Many knew the general shape but very few remembered any details so they had no idea which coordinates to draw their line or band through. Some gave no real thought to the problem and drew graphs that did not make sense. For example some graphs went vertical or turned back on themselves.

Part (c)(ii) was very discriminating. Less able candidates simply referred to any radiation that came to mind and forwarded very little explanation. The bulk of the candidates knew beta minus radiation was emitted but they were not careful how they expressed their reasons. For example, stating that there are more neutrons than protons is not sufficient to imply the nuclei are neutron rich. A majority in this group of candidates made no reference to the graph at all. Many that did had flawed reasoning. They thought the isotopes were neutron rich because the large number of free neutrons in the core. Good candidates also got into some difficulty by not reading the question carefully. Typically these candidates would start their answer with, 'If there are a lot more neutrons than protons then... but if the neutron to proton ratio is small then…'. These candidates obviously knew the subject matter but did not score many marks as the question clearly asks for a choice to be made.

E3. Even though part (a) needed a little thought almost all students obtained the correct answer. By contrast part (b)(i) was simply a factual recall question, which was answered poorly by a significant minority. The main error was for students not to state the energy needs to be given out or is required, when a nucleus was formed or broken up. It was common to see written, ‘The energy to keep the nucleus together’. In part (b)(ii) a majority of students simply read the value from the graph and gave an answer near 7.88 MeV without appreciating the ‘per nucleon’ on the y-axis of the graph. Part (c)(i) was done well by most students. Some students missed marks due to a lack of care in choosing specific coordinates for the graphs to pass through. Most students made a good attempt at part (c)(ii). Part (c)(iii) was more difficult and only the better student could correctly combine the two equations required to answer the question. A common mistake made by a few students who looked as if they were going to get the correct answer was for them to confuse the time units they were using. These students obtained the correct answer but then multiplied it by 60×60×24×365.

E4. In scripts where candidates showed no numerical values on the N axis, it was not


possible to judge the value of answers in part (a), so no marks could be awarded. The graph ought to start in an approximately linear fashion. To satisfy the requirements of the question, it should continue in an upward curve to meet Z = 80 at an N value around 120. The initial linearity was generally known, but the features of the upper part of the graph were less familiar to candidates. Some had no idea about how to respond, their answers bearing more resemblance to NIZ graphs showing α and β– decay chains.

Attempts at part (b) were usually more successful. Whatever mistakes had been made on the N/Z curve, examiners simply wanted candidates to show points that were respectively slightly below (at Z > 60), slightly above, and slightly below whatever line was drawn.

Part (c) attracted many highly successful solutions, with convincing (and often very different) explanations about why the loss of four a particles had to be accompanied by the emission of four β– particles in order to form the isotope 206Pb.

Candidates’ responses to part (d) were somewhat variable. It was necessary to identify two acceptable processes for the first mark; this was then followed by one mark each for the two discussions. Candidates who referred to β+ emission and to electron capture found it easier to score the discussion marks than those who chose a emission or proton emission. In the cases of β+ emission and electron capture, the conversion of a proton into a neutron produces an immediate and obvious change in the p/n ratio, a emission requires rather more careful explanation to indicate the effect on this ratio, whilst proton emission is very rare and occurs only to nuclides with sufficient instability for electrostatic repulsion to overcome the strong force.

E5.Part (a) was answered well but the last mark, which related path length to the energy of the ionising particle, was often missed. It was curious to note that candidates who gave the answer “beta particle” in part (a) (i) then, contrary to plain facts, often stated in part (a)(ii) “. . because there is a continuous range of path lengths... “. In part (b)(i) the equations showing alpha decay were easy marks for most candidates and the problem relating alpha particle energy to half-life only caused difficulties for the weaker candidates.

Most candidates failed to score in part (b)(iv) because it was necessary to state a little more than “because its neutron rich” to gain the mark. This was because it could be said that U-235 has a high neutron to proton ratio.


E7. (a) (i) The strong nuclear force is not well understood by many candidates. Few were able to state that it acts on all nucleons and therefore can be equal to electrostatic repulsion (ignoring gravitation) in the equilibrium position.

(ii) Although many candidates hinted that the presence of extra neutrons would separate the protons and thus reduce the electrostatic repulsion between them, few were able to state this unambiguously.

(iii) Most candidates making a reasonable attempt at this part failed to explain the dual nature of the strong nuclear force - that it is short ranged and attractive but becomes heavily repulsive at very short distances. Of those candidates stating either or both of these points, few were able to go on to consider the ramifications in terms of variable density if this force did not act in this way.

(b) (i) Many candidates correctly calculated the gravitational and electrostatic forces. Confusion between the forces and the field strengths was not infrequent as was the failure to square the radius. A minority of candidates were not able to

relate kto 1/4

(ii) Although most candidates appeared to recognise that the gravitational force was smaller than the electrostatic one, a significant number failed to state that it was so much smaller as to render it negligible.

· Web view2021. 3. 14. · 3.8.1.4 Nuclear Instability. Q1. A rod made from uranium−238 (U)...

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