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Revision 5 – Unit 5 – Chapter 9 - Hinchley Wood School

Q1.The high energy electron diffraction apparatus represented in Figure 1 can be used to determine nuclear radii. The intensity of the electron beam received by the detector is measured at various diffraction angles, θ.

Figure 1

(a) Sketch on the axes below a graph of the results expected from such an electron diffraction experiment.

(2)

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(b) (i) Use the data in the table to plot a straight line graph that confirms the relationship

elementradius of nucleus, R 10–15m

nucleonnumber; A

lead 6.66 208

tin 5.49 120

iron 4.35 56

silicon 3.43 28

carbon 2.66 12

(ii) Estimate the value of r0 from the graph.

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(c) Discuss the merits of using high energy electrons to determine nuclear radii rather than using α particles.

You may be awarded marks for the quality of written communication in your answer.

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(Total 10 marks)

Q2. (a) The lead nuclide Pb is unstable and decays in three stages through α and β

emissions to a different lead nuclide Pb. The position of these lead nuclides on a grid of neutron number, N, against proton number, Z, is shown below.

On the grid draw three arrows to represent one possible decay route.Label each arrow with the decay taking place.

(3)

(b) The copper nuclide Cu may decay by positron emission or by electron capture to form a nickel (Ni) nuclide.Complete the two equations that represent these two possible modes of decay.

positron emission Cu

electron capture Cu

(4)

(c) The nucleus of an atom may be investigated by scattering experiments in which radiation or particles bombard the nucleus.

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Name one type of radiation or particle that may be used in this investigation and describe the main physical principle of the scattering process.

State the information which can be obtained from the results of this scattering.


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(Total 10 marks)

Q3. The radioactive isotope of sodium has a half life of 2.6 years. A particular sample of this isotope has an initial activity of 5.5 × 105 Bq (disintegrations per second).

(a) Explain what is meant by the random nature of radioactive decay.

You may be awarded marks for the quality of written communication provided in your answer.

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(b) Use the axes to sketch a graph of the activity of the sample of sodium over a period of 6 years.

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(2)

(c) Calculate

(i) the decay constant, in s–1, of , 1 year = 3.15 × 107 s

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(ii) the number of atoms of in the sample initially,

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(iii) the time taken, in s, for the activity of the sample to fall from 1.0 × 105 Bq to 0.75 × 105 Bq.

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(Total 10 marks)

Q4. (a) Suggest two reasons why an α particle causes more ionisation than a β particle of the same initial kinetic energy.


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(b) A radioactive source has an activity of 3.2 × 109 Bq and emits α particles, each with kinetic energy of 5.2 Me V. The source is enclosed in a small aluminium container of mass 2.0 × 10–4 kg which absorbs the radiation completely.

(i) Calculate the energy, in J, absorbed from the source each second by the aluminium container.

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(ii) Estimate the temperature rise of the aluminium container in 1 minute, assuming no energy is lost from the aluminium.

specific heat capacity of aluminium = 900 J kg–1 K–1

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(Total 7 marks)

Q5. In an experiment to investigate the structure of the atom, α particles are directed normally at a thin metal foil which causes them to be scattered.

(a) (i) In which direction will the number of α particles per second be a maximum?

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(ii) State what this result suggests about the structure of the atoms in the metal.

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(b) A small number of α particles are scattered through 180°.

Explain what this suggests about the structure of the atoms in the metal.

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(c) The figure shows the path of an α particle passing near a nucleus.

(i) Name the force that is responsible for the deflection of the α particle.

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(ii) Draw an arrow on the diagram in the direction of the force on the α particle in the position where the force is a maximum.

(iii) The nucleus is replaced with one which has a larger mass number and a smaller proton number.

Draw on the diagram the path of an α particle that starts with the same velocity and position as that of the α particle drawn.

(4)(Total 8 marks)

Q6.The age of an ancient boat may be determined by comparing the radioactive decay of from living wood with that of wood taken from the ancient boat.A sample of 3.00 × l023 atoms of carbon is removed for investigation from a block of living

wood. In living wood one in 1012 of the carbon atoms is of the radioactive isotope , which has a decay constant of 3.84 × 10–12 s–1.

(a) What is meant by the decay constant?

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(b) Calculate the half-life of in years, giving your answer to an appropriate number of significant figures.

1 year = 3.15 × 107 s

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

(c) Show that the rate of decay of the atoms in the living wood sample is 1.15 Bq.

(2)

(d) A sample of 3.00 × 1023 atoms of carbon is removed from a piece of wood taken

from the ancient boat. The rate of decay due to the atoms in this sample is 0.65 Bq.Calculate the age of the ancient boat in years.

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

(e) Give two reasons why it is difficult to obtain a reliable age of the ancient boat from the carbon dating described.

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(Total 11 marks)

Q7. 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

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(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.

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

(Total 10 marks)

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M1.(a) graph to show: electron intensity decreasing with angle of diffraction (1) to a non-zero first minimum (1)

2

(b) (i) last column of table completed correctly (1)with either

A1/3

5.93

4.933.83

3.042.29

or

R3/(10 -45m3)295

16582.3

40.418.8

axes cover more than 50% of graph sheet (1)all points plotted correctly using labelled axes(i.e. x-axis A1/3, y-axis R/10–15m or x axis A, y-axis R3/10–45m3) (1)

(ii) gradient = r0 (1) [or gradient = r03]

gives r0 = (1.1 ± 0.1) × 10–15m (1)5

(c) electrons are not subject to the strong nuclear force (1)(so) electron scattering patterns are easier to interpret (1)electrons give greater resolution[or electrons are more accurate because they can get closer][or α particles cannot get so close to the nucleus because ofelectrostatic repulsion] (1)electrons give less recoil (1)(high energy) electrons are easier to produce[or electrons have a lower mass/ larger Q/m, so easier to accelerate] (1)

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(in Rutherford scattering) with α particles, the closest distanceof approach, not R is measured (1)

max 3QWC 1

[10]

M2. (a) (on grid: first arrow to start from ;

arrows must be consecutive;

last arrow must end on )

arrow showing the change for an α emission (1)arrow showing the change for a β emission (1)correct α and two β emissions in any order (1)

3

(b) (positron emission) → + β+ + ve (+Q) (1) (1)

(electron capture) + → + v(e) (+Q) (1) (1)4

(c) (the following examples may be included)

α particles (1)coulomb/electrostatic/electromagnetic repulsion[or K.E. converted to P.E. (as α particle approaches nucleus)] (1)information: any of the following: proton number, nuclear charge, upper limit to nuclear radius mass of nucleus is most of the mass of atom (1)

[alternative(high energy) electron (scattering) (1)diffraction of de Broglie Waves by nucleus (1)information: any of the following: nuclear radius, nuclear density (1)]

3QWC 2

[10]

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M3. (a) (use of ‘isotope’ instead of ‘nucleus’ not accepted)there is equal probability of any nucleus decaying,it cannot be known which particular nucleus will decay next,it cannot be known at what time a particular nucleus will decay,the rate of decay is unaffected by the surrounding conditions,it is only possible to estimate the proportion of nuclei decayingin the next time intervalany two statements (2)

2QWC 2

(b) continuous curve starting at 5.5 × 105 Bqplus correct 1st half-life (2.6 yrs, 2.75 × 105 Bq (1)correct 2nd half-life (5.2 years, 1.4 × 105 Bq) (1)(allow C.E. for incorrect 1st half-life)

2

(c) (i) (use of T1/2 = gives) λ = (1)

= 8.5 × 10–9 (s–1) (1)

(8.46 × 10–9 (s–1))

(ii) (use of = –λN gives) N = (1)

= 6.5 × 1013 (atoms) (1)

(allow C.E, for value of λ from (i))

(iii) (use of N = N0e–λt and A N gives)

= (1)

= 3.4 × 107 (s) (1)

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(allow C.E. for value of λ from (i))6

[10]

M4. (a) reasons:α particle has much more mass/momentum than β particleα particle has twice as much charge as a β particleα particle travels much slower than a β particle any two (1) (1)

2QWC 1

(b) (i) energy absorbed per sec (= energy released per sec)= 3.2 × 109 × 5.2 × 106 ×1.6 ×10–19 (1)= 2.7 ×10–3 (J) (1) (2.66 × 10–3 (J))

(ii) temperature rise in 1 minute

= (for numerator) (1) (for denominator) (1)

= 0.90 K (or °C) (1)

(allow C.E. for incorrect value in (i))5

[7]

M5. (a) (i) straight on or deflection of zero degrees (1)

(ii) the atom consists mainly of open space[or volume of nucleus is (very much) smaller than volume

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of the atom] (1)2

(b) most of the mass of an atom is contained in its nucleus[or the mass of the nucleus is greater than the mass of the α particle] (1)the nucleus contains a positive charge (1)the charge is concentrated at the nucleus (1)

max 2

(c) (i) electrostatic (force)[or electromagnetic or coulomb] (1)

(ii) arrow pointing away from the nucleus at the closest distance to the nucleus (1)

(iii) path showing less deflection at all times4

[8]

M6.(a) probability of decay per unit time/given time period

or fraction of atoms decaying per second

or the rate of radioactive decay is proportional to the number of (unstable)nuclei

and nuclear decay constant is the constant of proportionality (1)1

(b) use of =

= ln2/3.84 × 10–12 s (1) (1.805 × 1011 s)= (1.805 × 1011/3.15 × 107) = 5730 y (1)

answer given to 3 sf (1)3

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(c) number of nuclei = N = 3.00 × 1023 × 1/1012 (1)

(= 3.00 × 1011 nuclei)

(using )

rate of decay = 3.84 × 10–12 × 3.00 × 1011 (1)

(= 1.15 Bq)2

(d) (N = N0e–λt and activity is proportional to the number of nuclei A N use ofA = A0e–λt)

0.65 = 1.15 × (1)

t = 4720 y (1)3

(e) the boat may have been made with the wood some time after the tree wascut down

the background activity is high compared to the observed count rates

the count rates are low or sample size/mass is small or there is statisticalvariation in the recorded results

possible contamination

uncertainty in the ratio of carbon-14 in carbon thousands of years agoany two (1)(1)

2[11]

M7. (a)

β = 6 1

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(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]

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E1.Many candidates had no real idea about the shape of the graph expected in part (a). Straight lines through the origin - obviously an outright guess - were not uncommon. Amongst the better efforts which were not fully correct were the single slit diffraction pattern for light, showing several minima (instead of just one), those minima having zero intensity. A small minority failed to see that, on Figure 1, θ was shown as the angle between the deviated electron beam and the incident beam; these candidates drew a symmetrical graph which included the mirror image of the shape expected.

Answers to part (b), the graphical estimate of ro, were usually the most rewarding part of this question. Most candidates chose to plot R against A1/3, which gave better-spaced points than the alternative plot of R3 against A. A common failing amongst those who chose the latter route was to assume that the gradient would be ro, instead of ro

3. Graphical work was generally good, with well-chosen scales, five nicely plotted points and an acceptable straight line, but the choice of multiples of 3 for the major scale divisions made plotting a challenging task for some. In an occasional script, even when an acceptable graph had been drawn, an attempt would be made to calculate ro from an individual result instead of by use of the gradient. Since the question required the result to be found from the graph, this was not acceptable. The most serious misconception in part (b) was, when calculating the data, to assume that A1/3 stands for (A ÷ 3). An appreciable number of candidates made this error.

The mark scheme adopted for part (c) meant that those candidates who had any familiarity with high energy electron scattering were easily able to reach the maximum of three marks. Most answers were, however, very disappointing. The overall impression was that many candidates were unfamiliar with techniques involved in determining the size of the nucleus, and that some were relying on the knowledge of á particle scattering gained from GCSE. The main point in the answer should have been that electrons, unlike á particles, are not subject to the strong nuclear force. Substantial numbers knew this, but many fewer went on to indicate that it makes electron scattering patterns easier to interpret. References to electrostatic forces were frequent, but some candidates stated that electrons would be repelled (or á particles electrically attracted) by the nucleus. Another common misconception was that the relatively large mass of the á particle would readily cause it to have a short de Broglie wavelength. Overall, simplistic statements such as “electrons are very small and can get through tiny spaces in the target” were all too prevalent.

E2. The majority of candidates understood the requirements of part (a) and were able to give three arrows as a single route, but a significant number did not know the correct direction of both types of decay. Horizontal and vertical arrows were commonly seen. A common mistake was an apparent interpretation of the N on the vertical axis to mean nucleon number instead of neutron number, as the question stated. Several candidates

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had very little idea how to proceed and drew random lines which finished well away from the daughter nucleus.

Part (b), where two equations had to be completed was not well done with few candidates scoring full marks and many scoring zero. The Ni nuclide often appeared as Cu, x or y, frequently with incorrect values in the subscript and/or superscript and occasionally with different values in the decay mode. Weaker candidates often put the electron on the wrong side of the decay equation. Neutrinos appeared in most answers, but most frequently as a neutrino in one decay mode and an antineutrino in the other.

α particles were the choice of the great majority of candidates for the scattering particles in part (c) and for many candidates this was their only mark for this section. For most candidates, the requirement to describe the main physical principle of the scattering process was interpreted as ‘write as much as you can about Rutherford’s scattering experiment’, resulting in answers which were imprecise and demonstrated little more knowledge or understanding than that covered at GCSE. The idea that the experiment gave an upper limit for the nuclear radius was missed by nearly all candidates. The small minority of candidates who described electron diffraction by the nucleus usually answered the question well, demonstrating appropriate knowledge about the intensity minimum and the information gained from it. These candidates seemed to have a deeper understanding of the experiment, perhaps due to the fact that this was a new situation which had been taught in more detail at this level.

E3. The question on nuclear instability performed well on the whole and many candidates gained high marks. Answers to part (a) however, were weak, with most candidates achieving only one of the two available marks. This was disappointing since the examiners had recognised as many as five possible marking points. The reason for not gaining the two marks was either because the candidates elaborated unnecessarily on what was essentially one point or because of an imprecise use of language. Many candidates used ‘randomly’ in the context of their answer, without really explaining its meaning. Since the same word was part of the subject of the question, credit was not awarded for such answers. Most candidates gained credit by referring to the uncertainty in knowing which nucleus would decay or when. The examiners did not accept statements which referred to an isotope rather than a nucleus.

The graph in part (b) was usually drawn correctly with three points on it: the first at time zero and the other two at each subsequent half-life.

The calculation in part (c) was well done by most candidates, with many scoring full or almost full marks. A significant number however, were unable to progress beyond the

exponential decay equation in part (iii) and many had in .

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E4. Many answers in part (a) lacked sufficient detail to gain more than one or two marks. Although candidates knew that the mass of an α particle was greater than that of a β particle, few stated that its mass or momentum was much greater. Again, candidates knew that an α particle had more charge than a β particle but few stated it had twice as much charge. Few candidates realised that an α particle travels much slower than a β particle with the same kinetic energy; some even claimed that it travelled faster because it had more momentum. In addition, candidates often wrote at length about the relative penetrating powers of the two types of radiation.

In part (b) (i), many candidates gave a correct calculation although some candidates made arithmetical errors, often in the conversion of MeV to J, or in rounding the answer incorrectly. In part (ii), the underlying principles behind the calculation were known but some candidates failed to realise that the increase in temperature in one minute was required.

E5. Part (a) was answered well by most candidates but in part (b) only the better ones produced good answers. These candidates appreciated that in order for backscattering to occur, the á particle must collide with a particle of greater mass. The fact that repulsion came from having both a positive nucleus and a positive á particle was widely known.

Part (c) proved to be a good discriminator but it was common to see the force, which was supposed to act on the á particle, appearing to act on the nucleus.

E6.Less than half the candidates could explain the meaning of the decay constant. By contrast almost all candidates could find the half-life in part (b) and a majority could answer part (c). Some candidates did not gain credit because they conveniently removed 1012 in their calculation without showing the division. So lines like, 1.15 × 1012 Bq = 1.15 Bq, were seen. Most candidates who tackled part (d) using the exponential decay of the activity equation got full marks. Only a few candidates could not rearrange the equation. By contrast almost all candidates who tried to use the exponential decay in the number of nuclei got confused. Most had numbers of nuclei on one side of the equation but activity on the other.

Part (e) did discriminate but only between scoring zero marks or one mark. Very few

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candidates attempted two reasons. Most acceptable answers to this question were difficult for the candidate to express. For example, in question (d) it states that the decay rate due to carbon-14 is 0.65 Bq, indicating it is a corrected count rate. So an answer to part (e) like, ‘the background can effect the result’, is not acceptable. This is not the same as saying it is difficult to obtain the results for the sample activity because the background activity is high in comparison. This example is also ambiguous in that it suggests the surroundings can influence the rate of decay. Another answer that was not acceptable was, ‘radioactive decay is random so it’s bound to give false values’. To gain a mark following this line of thought it was necessary to refer to its effect on the statistics. The most common answers that candidates found easy to express included the following; the tree died well before the boat was made; or the boat was repaired later in its life with fresh wood; or that carbon based microbes died in the wood when the boat was rotting at the end of its useful life.

E7. 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.

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