topic 7 revision [143 marks]...topic 7 revision [143 marks]1. which feynman diagram shows beta-plus...
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Topic 7 Revision [143 marks]
1. Which Feynman diagram shows beta-plus (β ) decay?
MarkschemeA
Examiners report[N/A]
+
2. The average binding energy per nucleon of the nucleus is 7.5 MeV. What is thetotal energy required to separate the nucleons of one nucleus of ?
A. 53 MeV
B. 60 MeV
C. 113 MeV
D. 173 MeV
MarkschemeC
Examiners report[N/A]
158 O
158 O
Two pure samples of radioactive nuclides X and Y have the same initial number of
[1 mark]
[1 mark]
3. Two pure samples of radioactive nuclides X and Y have the same initial number ofatoms. The half-life of X is .
After a time equal to 4 half-lives of X the ratio is .
What is the half-life of Y?
A.
B.
C.
D.
MarkschemeD
Examiners report[N/A]
T 12
number of atoms of Xnumber of atoms of Y
18
0.25T 12
0.5T 12
3T 12
4T 12
4. The energy-level diagram for an atom that has four energy states is shown.
What is the number of different wavelengths in the emission spectrum of this atom?
A. 1
B. 3
C. 6
D. 7
MarkschemeC
Examiners report[N/A]
A detector, placed close to a radioactive source, detects an activity of 260 Bq. The
[1 mark]
[1 mark]
5. A detector, placed close to a radioactive source, detects an activity of 260 Bq. Theaverage background activity at this location is 20 Bq. The radioactive nuclide has a half-life of 9 hours.
What activity is detected after 36 hours?
A. 15 Bq
B. 16 Bq
C. 20 Bq
D. 35 Bq
MarkschemeD
Examiners report[N/A]
6. Element X decays through a series of alpha (α) and beta minus (β ) emissions. Whichseries of emissions results in an isotope of X?
A. 1α and 2βB. 1α and 4βC. 2α and 2βD. 2α and 3β
MarkschemeA
Examiners report[N/A]
–
–
–
–
–
7. A graph of the variation of average binding energy per nucleon with nucleon numberhas a maximum. What is indicated by the region around the maximum?
A. The position below which radioactive decay cannot occur
B. The region in which fission is most likely to occur
C. The position where the most stable nuclides are found
D. The region in which fusion is most likely to occur
[1 mark]
[1 mark]
[1 mark]
MarkschemeC
Examiners report[N/A]
8. Three of the fundamental forces between particles are
I. strong nuclear
II. weak nuclear
III. electromagnetic.
What forces are experienced by an electron?
A. I and II only
B. I and III only
C. II and III only
D. I, II and III
MarkschemeC
Examiners report[N/A]
9. What is correct about the Higgs Boson?
A. It was predicted before it was observed.
B. It was difficult to detect because it is charged.
C. It is not part of the Standard Model.
D. It was difficult to detect because it has no mass.
MarkschemeA
[1 mark]
[1 mark]
Examiners report[N/A]
10. Identify the conservation law violated in the proposed reaction.
p + p → p + n + µ
A. Strangeness
B. Lepton number
C. Charge
D. Baryon number
MarkschemeB
Examiners report[N/A]
+ + + 0 +
11a.
The radioactive nuclide beryllium-10 (Be-10) undergoes beta minus (β–) decay to form a stableboron (B) nuclide.
Identify the missing information for this decay.
Markscheme
conservation of mass number AND charge ,
Correct identification of both missing values required for [1].
[1 mark]
Examiners report[N/A]
104 Be →10
5 B + β + ¯¯Ve
105 B 10
4 Be
[1 mark]
[1 mark]
11b.
The initial number of nuclei in a pure sample of beryllium-10 is N . The graph shows how thenumber of remaining beryllium nuclei in the sample varies with time.
On the graph, sketch how the number of boron nuclei in the sample varies with time.
Markschemecorrect shape ie increasing from 0 to about 0.80 N
crosses given line at 0.50 N
[2 marks]
Examiners report[N/A]
0
0
0
After 4.3 × 10 years,6
[2 marks]
11c. After 4.3 × 10 years,
Show that the half-life of beryllium-10 is 1.4 × 10 years.
MarkschemeALTERNATIVE 1
fraction of Be = , 12.5%, or 0.125
therefore 3 half lives have elapsed
«≈ 1.4 × 10 » «y»
ALTERNATIVE 2
fraction of Be = , 12.5%, or 0.125
leading to λ = 4.836 × 10 «y»
= 1.43 × 10 «y»
Must see at least one extra sig fig in final answer.
[3 marks]
Examiners report[N/A]
6
= 7.number of produced boron nuclei
number of remaining beryllium nuclei
6
18
t = = 1.43 × 10612
4.3×106
36
18
= e−λ(4.3 × 106)18
–7 –1
ln2λ
6
Beryllium-10 is used to investigate ice samples from Antarctica. A sample of ice initially
[3 marks]
11d. Beryllium-10 is used to investigate ice samples from Antarctica. A sample of ice initiallycontains 7.6 × 10 atoms of beryllium-10. State the number of remaining beryllium-10nuclei in the sample after 2.8 × 10 years.
Markscheme1.9 × 10
[1 mark]
Examiners report[N/A]
11
6
11
11e.
An ice sample is moved to a laboratory for analysis. The temperature of the sample is –20 °C.
State what is meant by thermal radiation.
Markschemeemission of (infrared) electromagnetic/infrared energy/waves/radiation.
[1 mark]
Examiners report[N/A]
Discuss how the frequency of the radiation emitted by a black body can be used to
[1 mark]
[1 mark]
11f. Discuss how the frequency of the radiation emitted by a black body can be used toestimate the temperature of the body.
Markschemethe (peak) wavelength of emitted em waves depends on temperature of emitter/reference toWein’s Law
so frequency/color depends on temperature
[2 marks]
Examiners report[N/A]
11g. Calculate the peak wavelength in the intensity of the radiation emitted by the icesample.
Markscheme
= 1.1 × 10 «m»
Allow ECF from MP1 (incorrect temperature).
[2 marks]
λ = 2.90×10−3
253
–5
[2 marks]
[2 marks]
Examiners report[N/A]
11h. Derive the units of intensity in terms of fundamental SI units.
Markschemecorrect units for Intensity (allow W, Nms OR Js in numerator)
rearrangement into proper SI units = kgs
Allow ECF for MP2 if final answer is in fundamental units.
[2 marks]
Examiners report[N/A]
–1 –1
–3
12a. Rutherford constructed a model of the atom based on the results of the alphaparticle scattering experiment. Describe this model.
[2 marks]
[2 marks]
Markscheme«most of» the mass of the atom is confined within a very small volume/nucleus
«all» the positive charge is confined within a very small volume/nucleus
electrons orbit the nucleus «in circular orbits»
[2 marks]
Examiners report[N/A]
12b.
Rhodium-106 ( ) decays into palladium-106 ( ) by beta minus (β ) decay.
The binding energy per nucleon of rhodium is 8.521 MeV and that of palladium is 8.550 MeV.
State what is meant by the binding energy of a nucleus.
Markschemethe energy needed to separate the nucleons of a nucleus
OR
energy released when a nucleus is formed from its nucleons
Allow neutrons AND protons for nucleons
Don’t allow constituent parts
[1 mark]
Examiners report[N/A]
10645Rh 106
46Pd –
Show that the energy released in the β decay of rhodium is about 3 MeV.–
[1 mark]
12c. Show that the energy released in the β decay of rhodium is about 3 MeV.
MarkschemeQ = 106 × 8.550 − 106 × 8.521 = 3.07 «MeV»
«Q ≈ 3 Me V»
[1 mark]
Examiners report[N/A]
–
12d.
β decay is described by the following incomplete Feynman diagram.
Draw a labelled arrow to complete the Feynman diagram.
–
[1 mark]
[1 mark]
Markschemeline with arrow as shown labelled anti-neutrino/
Correct direction of the “arrow” is essential
The line drawn must be “upwards” from the vertex in the time direction i.e. above thehorizontal
[1 mark]
Examiners report[N/A]
v
12e. Identify particle V.
MarkschemeV = W
Examiners report[N/A]
–
[1 mark]
13a. Identify the missing information for this decay.
[1 mark]
[2 marks]
Markscheme
antineutrino AND charge AND mass number of electron ,
conservation of mass number AND charge ,
Do not accept V.
Accept without subscript e.
[2 marks]
Examiners report[N/A]
104 Be →10
5 B + 0−1 e + ¯¯Ve
0−1e ¯¯V
105 B 10
4 Be
V
13b. On the graph, sketch how the number of boron nuclei in the sample varies with time.
Markschemecorrect shape ie increasing from 0 to about 0.80 N
crosses given line at 0.50 N
[2 marks]
Examiners report[N/A]
0
0
After 4.3 × 10 years,6
[2 marks]
13c. After 4.3 × 10 years,
Show that the half-life of beryllium-10 is 1.4 × 10 years.
MarkschemeALTERNATIVE 1
fraction of Be = , 12.5%, or 0.125
therefore 3 half lives have elapsed
«≈ 1.4 × 10 » «y»
ALTERNATIVE 2
fraction of Be = , 12.5%, or 0.125
leading to λ = 4.836 × 10 «y»
= 1.43 × 10 «y»
Must see at least one extra sig fig in final answer.
[3 marks]
Examiners report[N/A]
6
= 7.number of produced boron nuclei
number of remaining beryllium nuclei
6
18
t = = 1.43 × 10612
4.3×106
36
18
= e−λ(4.3 × 106)18
–7 –1
ln2λ
6
Beryllium-10 is used to investigate ice samples from Antarctica. A sample of ice
[3 marks]
13d. Beryllium-10 is used to investigate ice samples from Antarctica. A sample of iceinitially contains 7.6 × 10 atoms of beryllium-10. The present activity of the sample is8.0 × 10 Bq.
Determine, in years, the age of the sample.
Markschemeλ «=
» = 4.95 × 10 «y »
rearranging of A = λN e to give –λt = ln «= –0.400»
t = «y»
Allow ECF from MP1
[3 marks]
Examiners report[N/A]
11
−3
ln21.4×106
–7 –1
0–λt 8.0×10−3×365×24×60×60
4.95×10−7×7.6×1011
= 8.1 × 105−0.400−4.95×10−7
13e. State what is meant by thermal radiation.
Markschemeemission of (infrared) electromagnetic/infrared energy/waves/radiation.
[1 mark]
[3 marks]
[1 mark]
Examiners report[N/A]
13f. Discuss how the frequency of the radiation emitted by a black body can be used toestimate the temperature of the body.
Markschemethe (peak) wavelength of emitted em waves depends on temperature of emitter/reference toWein’s Law
so frequency/color depends on temperature
[2 marks]
Examiners report[N/A]
13g. Calculate the peak wavelength in the intensity of the radiation emitted by the icesample.
[2 marks]
[2 marks]
Markscheme
= 1.1 × 10 «m»
Allow ECF from MP1 (incorrect temperature).
[2 marks]
Examiners report[N/A]
λ = 2.90×10−3
253
–5
13h. The temperature in the laboratory is higher than the temperature of the ice sample.Describe one other energy transfer that occurs between the ice sample and thelaboratory.
Markschemefrom the laboratory to the sample
conduction – contact between ice and lab surface.
OR
convection – movement of air currents
Must clearly see direction of energy transfer for MP1.
Must see more than just words “conduction” or “convection” for MP2.
[2 marks]
Examiners report[N/A]
[2 marks]
14a. Bohr modified the Rutherford model by introducing the condition mvr = n . Outlinethe reason for this modification.
Markschemethe electrons accelerate and so radiate energy
they would therefore spiral into the nucleus/atoms would be unstable
electrons have discrete/only certain energy levels
the only orbits where electrons do not radiate are those that satisfy the Bohr condition «mvr= n
»
[3 marks]
Examiners report[N/A]
h2π
h2π
14b. Show that the speed v of an electron in the hydrogen atom is related to the radius r ofthe orbit by the expression
where k is the Coulomb constant.
v = √ ke2
mer
[3 marks]
[1 mark]
Markscheme
OR
KE = PE hence m v =
«solving for v to get answer»
Answer given – look for correct working
[1 mark]
Examiners report[N/A]
=mev2
rke2
r2
12
12 e
2 12
ke2
r
14c. Using the answer in (b) and (c)(i), deduce that the radius r of the electron’s orbit in theground state of hydrogen is given by the following expression.
r =h2
4π2kmee2
[2 marks]
Markschemecombining v = with m vr = using correct substitution
«eg»
correct algebraic manipulation to gain the answer
Answer given – look for correct working
Do not allow a bald statement of the answer for MP2. Some further working eg cancellationof m or r must be shown
[2 marks]
Examiners report[N/A]
√ ke2
mer eh
2π
me2 r2 =ke2
merh2
4π2
14d. Calculate the electron’s orbital radius in (c)(ii).
Markscheme« r =
»
r = 5.3 × 10 «m»
[1 mark]
Examiners report[N/A]
(6.63×10−34)2
4π2×8.99×109×9.11×10−31×(1.6×10−19)2
–11
106Rh 106Pd –
[1 mark]
14e.
Rhodium-106 ( ) decays into palladium-106 ( ) by beta minus (β ) decay. Thediagram shows some of the nuclear energy levels of rhodium-106 and palladium-106. The arrowrepresents the β decay.
Explain what may be deduced about the energy of the electron in the β decay.
Markschemethe energy released is 3.54 – 0.48 = 3.06 «MeV»
this is shared by the electron and the antineutrino
so the electron’s energy varies from 0 to 3.06 «MeV»
[3 marks]
Examiners report[N/A]
10645Rh 106
46Pd –
–
–
14f. Suggest why the β decay is followed by the emission of a gamma ray photon.–
[3 marks]
[1 mark]
Markschemethe palladium nucleus emits the photon when it decays into the ground state «from theexcited state»
[1 mark]
Examiners report[N/A]
14g. Calculate the wavelength of the gamma ray photon in (d)(ii).
MarkschemePhoton energy
E = 0.48 × 10 × 1.6 × 10 = «7.68 × 10 J»
λ = « =» 2.6 × 10 «m»
Award [2] for a bald correct answer
Allow ECF from incorrect energy
[2 marks]
Examiners report[N/A]
6 –19 –14
=hcE
6.63×10−34×3×108
7.68×10−14–12
A positive pion decays into a positive muon and a neutrino.
[2 marks]
15a.
A positive pion decays into a positive muon and a neutrino.
The momentum of the muon is measured to be 29.8 MeV c in a laboratory reference frame inwhich the pion is at rest. The rest mass of the muon is 105.7 MeV c and the mass of theneutrino can be assumed to be zero.
For the laboratory reference frame
write down the momentum of the neutrino.
Markscheme«–»29.8 «MeVc »
[1 mark]
Examiners report[N/A]
π+ → μ+ + vμ
–1
–2
–1
15b. show that the energy of the pion is about 140 MeV.
[1 mark]
[2 marks]
MarkschemeE = + p c OR E = 109.8 «MeV»
E = « + 29.8 =» 139.6 «MeV»
Final value to at least 3 sig figs required for mark.
[2 marks]
Examiners report[N/A]
π √p2μc2 + m2
μc4 v µ
π√29.82 + 105.72
15c. State the rest mass of the pion with an appropriate unit.
Markscheme139.6 MeVc
Units required.
Accept 140 MeVc .
[1 mark]
Examiners report[N/A]
–2
–2
16. Which statement about atomic spectra is not true?
A. They provide evidence for discrete energy levels in atoms.
B. Emission and absorption lines of equal frequency correspond to transitions betweenthe same two energy levels.
C. Absorption lines arise when electrons gain energy.
D. Emission lines always correspond to the visible part of the electromagnetic spectrum.
[1 mark]
[1 mark]
MarkschemeD
Examiners report[N/A]
17. What gives the total change in nuclear mass and the change in nuclear binding energyas a result of a nuclear fusion reaction?
MarkschemeB
Examiners report[N/A]
18. The Feynman diagram shows a particle interaction involving a W boson.
Which particles are interacting?
A. U and Y
B. W boson and Y
C. X and Y
D. U and X
–
–
[1 mark]
[1 mark]
MarkschemeC
Examiners report[N/A]
19a.
The Feynman diagram shows electron capture.
Deduce that X must be an electron neutrino.
Markschemeit has a lepton number of 1 «as lepton number is conserved»
it has a charge of zero/is neutral «as charge is conserved»
OR
it has a baryon number of 0 «as baryon number is conserved»
Do not credit answers referring to energy
Examiners report[N/A]
[2 marks]
19b. Distinguish between hadrons and leptons.
Markschemehadrons experience strong force
OR
leptons do not experience the strong force
hadrons made of quarks/not fundamental
OR
leptons are not made of quarks/are fundamental
hadrons decay «eventually» into protons
OR
leptons do not decay into protons
Accept leptons experience the weak force
Allow “interaction” for “force”
Examiners report[N/A]
20a. State and explain the nature of the particle labelled X.
[2 marks]
[3 marks]
Markscheme«electron» neutrino
it has a lepton number of 1 «as lepton number is conserved»
it has a charge of zero/is neutral «as charge is conserved»ORit has a baryon number of 0 «as baryon number is conserved»
Do not allow antineutrino
Do not credit answers referring to energy
Examiners report[N/A]
20b.
Particles can be used in scattering experiments to estimate nuclear sizes.
Outline how these experiments are carried out.
Markscheme«high energy particles incident on» thin sample
detect angle/position of deflected particles
reference to interference/diffraction/minimum/maximum/numbers of particles
Allow “foil” instead of thin
Examiners report[N/A]
Outline why the particles must be accelerated to high energies in
[2 marks]
20c. Outline why the particles must be accelerated to high energies inscattering experiments.
Markschemeλ OR λ
so high energy gives small λto match the small nuclear size
Alternative 2
E = hf/energy is proportional to frequency
frequency is inversely proportional to wavelength/ c = fλto match the small nuclear size
Alternative 3
higher energy means closer approach to nucleus
to overcome the repulsive force from the nucleus
so greater precision in measurement of the size of the nucleus
Accept inversely proportional
Only allow marks awarded from one alternative
Examiners report[N/A]
∝ 1√E
∝ 1E
State and explain one example of a scientific analogy.
[3 marks]
20d. State and explain one example of a scientific analogy.
Markschemetwo analogous situations stated
one element of the analogy equated to an element of physics
eg: moving away from Earth is like climbing a hill where the contours correspond to theequipotentials
Atoms in an ideal gas behave like pool balls
The forces between them only act during collisions
Examiners report[N/A]
20e.
Electron diffraction experiments indicate that the nuclear radius of carbon-12 is 2.7 x 10 m.The graph shows the variation of nuclear radius with nucleon number. The nuclear radius of thecarbon-12 is shown on the graph.
Plot the position of magnesium-24 on the graph.
–15
[2 marks]
[1 mark]
Markschemecorrectly plotted
Allow ECF from (d)(i)
Examiners report[N/A]
20f. Draw a line on the graph, to show the variation of nuclear radius with nucleon number.
Markschemesingle smooth curve passing through both points with decreasing gradient
through origin
Examiners report[N/A]
Two pulses are travelling towards each other.
[2 marks]
21. Two pulses are travelling towards each other.
What is a possible pulse shape when the pulses overlap?
MarkschemeA
Examiners report[N/A]
22. A nucleus of phosphorus (P) decays to a nucleus of silicon (Si) with the emission ofparticle X and particle Y.
What are X and Y?
MarkschemeD
3015P → 30
14Si + X + Y
[1 mark]
[1 mark]
Examiners report[N/A]
23. What is the definition of the unified atomic mass unit?
A. the mass of a neutral atom of carbon-12
B. The mass of a neutral atom of hydrogen-1
C. the mass of a nucleus of carbon-12
D. The mass of a nucleus of hydrogen-1
MarkschemeA
Examiners report[N/A]
112
112
24. In nuclear fission, a nucleus of element X absorbs a neutron (n) to give a nucleus ofelement Y and a nucleus of element Z.
X + n → Y + Z + 2n
What is and ?
MarkschemeA
Examiners report[N/A]
magnitude of the binding energy per nucleon of Ymagnitude of the binding energy per nucleon of X
total binding energy of Y and Ztotal binding energy of X
What is the energy equivalent to the mass of one proton?
[1 mark]
[1 mark]
25. What is the energy equivalent to the mass of one proton?
A. 9.38 × (3 × 10 ) × 10 J
B. 9.38 × (3 × 10 ) × 1.6 × 10 J
C. J
D. 9.38 × 10 × 1.6 × 10 J
MarkschemeD
Examiners report[N/A]
8 2 6
8 2 –19
9.38×108
1.6×10−19
8 –19
26. Atomic spectra are caused when a certain particle makes transitions between energylevels.What is this particle?
A. Electron
B. Proton
C. Neutron
D. Alpha particle
MarkschemeA
Examiners report[N/A]
27. The half-life of a radioactive element is 5.0 days. A freshly-prepared sample contains128 g of this element. After how many days will there be 16 g of this element left behindin the sample?
A. 5.0 days
B. 10 days
C. 15 days
D. 20 days
[1 mark]
[1 mark]
[1 mark]
MarkschemeC
Examiners report[N/A]
28. The binding energy per nucleon of is 6 MeV. What is the energy required toseparate the nucleons of this nucleus?
A. 24 MeV
B. 42 MeV
C. 66 MeV
D. 90 MeV
MarkschemeC
Examiners report[N/A]
114 Be
29. The reaction p + n → p + does not occur because it violates the conservation lawof
A. electric charge.
B. baryon number.
C. lepton number.
D. strangeness.
MarkschemeB
Examiners report[N/A]
+ 0 + π0
A pure sample of nuclide A and a pure sample of nuclide B have the same activity at
[1 mark]
[1 mark]
30. A pure sample of nuclide A and a pure sample of nuclide B have the same activity attime t = 0. Nuclide A has a half-life of T, nuclide B has a half-life of 2T.
What is when t = 4T?
A. 4
B. 2
C.
D.
MarkschemeD
Examiners report[N/A]
activity of Aactivity of B
12
14
31. In the nuclear reaction X + Y → Z + W, involving nuclides X, Y, Z and W, energy isreleased. Which is correct about the masses (M) and the binding energies ( BE) of thenuclides?
MarkschemeC
Examiners report[N/A]
32. Which of the following leads to a paradigm shift?
A. Multi-loop circuits
B. Standing waves
C. Total internal reflection
D. Atomic spectra
[1 mark]
[1 mark]
[1 mark]
MarkschemeD
Examiners report[N/A]
33a. State the quark structures of a meson and a baryon.
MarkschemeMeson: quark-antiquark pairBaryon: 3 quarks
Examiners report[N/A]
Λ0
[2 marks]
33b.
A possible decay of a lambda particle ( ) is shown by the Feynman diagram.
Explain which interaction is responsible for this decay.
MarkschemeAlternative 1
strange quark changes «flavour» to an up quark
changes in quarks/strangeness happen only by the weak interaction
Alternative 2
Strangeness is not conserved in this decay «because the strange quark changes to an upquark»
Strangeness is not conserved during the weak interaction
Do not allow a bald answer of weak interaction.
Examiners report[N/A]
Λ0
33c. Draw arrow heads on the lines representing and d in the .u π−
[2 marks]
[1 mark]
Markschemearrows drawn in the direction shown
Both needed for [1] mark.
Examiners report[N/A]
33d. Identify the exchange particle in this decay.
MarkschemeW
Do not allow W or W .
Examiners report[N/A]
−
+
33e. Outline one benefit of international cooperation in the construction or use of high-energy particle accelerators.
[1 mark]
[1 mark]
Markschemeit lowers the cost to individual nations, as the costs are shared
international co-operation leads to international understanding OR historical example of co-operation OR co-operation always allows science to proceed
large quantities of data are produced that are more than one institution/research group canhandle co-operation allows effective analysis
Any one.
Examiners report[N/A]
34a.
The first scientists to identify alpha particles by a direct method were Rutherford and Royds.They knew that radium-226 ( ) decays by alpha emission to form a nuclide known as radon(Rn).
Write down the missing values in the nuclear equation for this decay.
Markscheme222 AND 4
Both needed.
Examiners report[N/A]
22686 Ra
Rutherford and Royds put some pure radium-226 in a small closed cylinder A. Cylinder
[1 mark]
34b. Rutherford and Royds put some pure radium-226 in a small closed cylinder A. CylinderA is fixed in the centre of a larger closed cylinder B.
At the start of the experiment all the air was removed from cylinder B. The alphaparticles combined with electrons as they moved through the wall of cylinder A to form heliumgas in cylinder B.
The wall of cylinder A is made from glass. Outline why this glass wall had to be very thin.
Markschemealpha particles highly ionizingORalpha particles have a low penetration powerORthin glass increases probability of alpha crossing glassORdecreases probability of alpha striking atom/nucleus/molecule
Examiners report[N/A]
Rutherford and Royds expected 2.7 x 10 alpha particles to be emitted during15
[1 mark]
34c. Rutherford and Royds expected 2.7 x 10 alpha particles to be emitted duringthe experiment. The experiment was carried out at a temperature of 18 °C. Thevolume of cylinder B was 1.3 x 10 m and the volume of cylinder A was negligible. Calculatethe pressure of the helium gas that was collected in cylinder B.
Markschemeconversion of temperature to 291 K
p = 4.5 x 10 x 8.31 x « »
OR
p = 2.7 x 10 x 1.38 x 10 x « »
0.83 or 0.84 «Pa»
Examiners report[N/A]
15
–5 3
–9 2.911.3×10−5
15 –23 2.911.3×10−5
Rutherford and Royds identified the helium gas in cylinder B by observing its emission
[3 marks]
34d. Rutherford and Royds identified the helium gas in cylinder B by observing its emissionspectrum. Outline, with reference to atomic energy levels, how an emission spectrumis formed.
Markschemeelectron/atom drops from high energy state/level to low state
energy levels are discrete
wavelength/frequency of photon is related to energy change or quotes E = hf or E =
and is therefore also discrete
Examiners report[N/A]
hcλ
34e. The work was first reported in a peer-reviewed scientific journal. Outlinewhy Rutherford and Royds chose to publish their work in this way.
[3 marks]
[1 mark]
Markschemepeer review guarantees the validity of the workORmeans that readers have confidence in the validity of work
OWTTE
Examiners report[N/A]
35a.
The first scientists to identify alpha particles by a direct method were Rutherford and Royds.They knew that radium-226 ( ) decays by alpha emission to form a nuclide known as radon(Rn).
Write down the nuclear equation for this decay.
Markscheme
OR
These must be seen on the right-hand side of the equation.
Examiners report[N/A]
22686 Ra
42α
42He22286 Rn
At the start of the experiment, Rutherford and Royds put 6.2 x 10 mol of pure radium-226 in a–4
[2 marks]
35b.
At the start of the experiment, Rutherford and Royds put 6.2 x 10 mol of pure radium-226 in asmall closed cylinder A. Cylinder A is fixed in the centre of a larger closed cylinder B.
The experiment lasted for 6 days. The decay constant of radium-226 is 1.4 x 10 s .
Deduce that the activity of the radium-226 is almost constant during the experiment.
–4
–11 –1
[2 marks]
MarkschemeALTERNATIVE 1
6 days is 5.18 x 10 s
activity after 6 days is
OR
A = 0.9999927 A or 0.9999927 N
OR
states that index of e is so small that is ≈ 1
OR
A – A ≈ 10 «s »
ALTERNATIVE 2shows half-life of the order of 10 s or 5.0 x 10 s
converts this to year «1600 y» or days and states half-life much longer than experimentcompared to experiment
Award [1 max] if calculations/substitutions have numerical slips but would lead to correctdeduction.
eg: failure to convert 6 days to seconds but correct substitution into equation will give MP2.
Allow working in days, but for MP1 must see conversion of or half-life to day .
Examiners report[N/A]
5
A0e−1.4×10−11×5.8×105≈ A0
0 λ 0
A
A0
0–15 –1
11 10
λ –1
35c. Show that about 3 x 10 alpha particles are emitted by the radium-226 in 6 days.15 [3 marks]
MarkschemeALTERNATIVE 1
use of A = N
conversion to number of molecules = nN = 3.7 x 10
OR
initial activity = 5.2 x 10 «s »
number emitted = (6 x 24 x 3600) x 1.4 x 10 x 3.7 x 10 or 2.7 x 10 alpha particles
ALTERNATIVE 2use of N = N
N = n x N = 3.7 x 10
alpha particles emitted «= number of atoms disintegrated = N – N =» N or 2.7 x 10 alpha particles
Must see correct substitution or answer to 2+ sf for MP3
Examiners report[N/A]
λ 0
A20
9 –1
–11 20 15
0e−λt
0 A20
0 0(1 − e−λ×6×24×3600) 15
35d.
At the start of the experiment, all the air was removed from cylinder B. The alpha particlescombined with electrons as they moved through the wall of cylinder A to form helium gas incylinder B.
The wall of cylinder A is made from glass. Outline why this glass wall had to be verythin.
[1 mark]
Markschemealpha particles highly ionizingORalpha particles have a low penetration powerORthin glass increases probability of alpha crossing glassORdecreases probability of alpha striking atom/nucleus/molecule
Do not allow reference to tunnelling.
Examiners report[N/A]
35e. The experiment was carried out at a temperature of 18 °C. The volume of cylinder Bwas 1.3 x 10 m and the volume of cylinder A was negligible. Calculate the pressureof the helium gas that was collected in cylinder B over the 6 day period. Helium is a monatomicgas.
Markschemeconversion of temperature to 291 K
p = 4.5 x 10 x 8.31 x « »
OR
p = 2.7 x 10 x 1.3 x 10 x « »
0.83 or 0.84 «Pa»
Allow ECF for 2.7 x 10 from (b)(ii).
–5 3
–9 2911.3×10−5
15 –23 2911.3×10−5
15
[3 marks]
Examiners report[N/A]
36. Photons of energy 2.3eV are incident on a low-pressure vapour. The energy levels ofthe atoms in the vapour are shown
What energy transition will occur when a photon is absorbed by the vapour?
A. –3.9eV to –1.6eV
B. –1.6eV to 0eV
C. –1.6eV to –3.9eV
D. 0eV to –1.6eV
MarkschemeA
Examiners report[N/A]
37. When an alpha particle collides with a nucleus of nitrogen-14 , a nucleus X canbe produced together with a proton. What is X?
A.
B.
C.
D.
MarkschemeB
(147 N)
188 X178 X189 X179 X
[1 mark]
[1 mark]
Examiners report[N/A]
38. The mass defect for deuterium is 4×10 kg. What is the binding energy of deuterium?
A. 4×10 eV
B. 8×10 eV
C. 2×10 eV
D. 2×10 eV
MarkschemeC
Examiners report[N/A]
–30
–7
–2
6
12
39. As quarks separate from each other within a hadron, the interaction between thembecomes larger. What is the nature of this interaction?
A. ElectrostaticB. Gravitational C. Strong nuclear D. Weak nuclear
MarkschemeC
Examiners report[N/A]
40. Which of the following lists the particles emitted during radioactive decay in order ofincreasing ionizing power?
A. γ, β, αB. β, α, γ C. α, γ, βD. α, β, γ
[1 mark]
[1 mark]
[1 mark]
MarkschemeA
Examiners report[N/A]
41. What is the charge on an electron antineutrino and during what process is an electronantineutrino produced?
MarkschemeD
Examiners report[N/A]
42a. A particular K meson has a quark structure s. State the charge on this meson.
Markschemecharge: –1«e» or negative or K
Negative signs required.
u
−
[1 mark]
[1 mark]
Examiners report[N/A]
42b. The Feynman diagram shows the changes that occur during beta minus (β ) decay.
Label the diagram by inserting the four missing particle symbols.
Markscheme
correct symbols for both missing quarks
exchange particle and electron labelled W or W and e or eDo not allow W or e or β Allow β or β
Examiners report[N/A]
–
– –
+ + + –
Carbon-14 (C-14) is a radioactive isotope which undergoes beta minus (β ) decay to–
[2 marks]
42c. Carbon-14 (C-14) is a radioactive isotope which undergoes beta minus (β ) decay tothe stable isotope nitrogen-14 (N-14). Energy is released during this decay. Explainwhy the mass of a C-14 nucleus and the mass of a N-14 nucleus are slightly different eventhough they have the same nucleon number.
Markschemedecay products include an electron that has massOR products have energy that has a mass equivalentORmass/mass defect/binding energy converted to mass/energy of decay products
«so»
mass C-14 > mass N-14ORmass of n > mass of pORmass of d > mass of u
Accept reference to “lighter” and “heavier” in mass.Do not accept implied comparison, eg “C-14 has greater mass”. Comparison must beexplicit as stated in scheme.
Examiners report[N/A]
–
43a. A particular K meson has a quark structure s. State the charge, strangeness andbaryon number for this meson.
u
[2 marks]
[2 marks]
Markschemecharge: –1«e» or negative or K
strangeness: –1
baryon number: 0
Negative signs required.Award [2] for three correct answers, [1 max] for two correct answer and [0] for one correctanswer.
Examiners report[N/A]
−
43b. The Feynman diagram shows the changes that occur during beta minus (β ) decay.
Label the diagram by inserting the four missing particle symbols and the direction of the arrowsfor the decay particles.
Markscheme
correct symbols for both missing quarks
exchange particle and electron labelled W or W and e or e
Do not allow W or e or β . Allow β or β .
arrows for both electron and anti-neutrino correct
Allow ECF from previous marking point.
–
– –
+ + + –
[3 marks]
Examiners report[N/A]
43c. C-14 decay is used to estimate the age of an old dead tree. The activity of C-14 in thedead tree is determined to have fallen to 21% of its original value. C-14 has a half-lifeof 5700 years.
(i) Explain why the activity of C-14 in the dead tree decreases with time.
(ii) Calculate, in years, the age of the dead tree. Give your answer to an appropriate number ofsignificant figures.
[4 marks]
Printed for Jyvaskylan Lyseon lukio
© International Baccalaureate Organization 2019 International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®
Markschemeinumber of C-14 atoms/nuclei are decreasingORdecreasing activity proportional to number of C-14 atoms/nucleiORA = A e so A decreases as t increasesDo not allow “particles”Must see reference to atoms or nuclei or an equation, just “C-14 is decreasing” is notenough.
ii0.21 = (0.5)OR
n = 2.252 half-lives or t =1 2834 «y»Early rounding to 2.25 gives 12825 y
13000 y rounded correctly to two significant figures:Both needed; answer must be in year for MP3.Allow ECF from MP2.Award [3] for a bald correct answer.
Examiners report[N/A]
0–λt
n
0.21 = e−( )ln2×t
5700