12 Quantum and nuclear physics

12 Quantum and nuclear physics

12.2 Nuclear physics

©OxfordUniversityPress2016www.oxfordsecondary.co.uk/acknowledgements

1

Nuclear radius, density and energy levels - answers

Nuclear radius and density

1 4 × 10−14 m

2 2.8 × 10−14 m

3 radius = 2.9 × 10−15 m

density = 2.3 × 1017 kg m−3

4 1.9 × 10−15 m

Nuclear energy levels

5 The excited aluminium nucleus emits gamma photons in order to return to the ground state. Either a single photon of energy 1.02 MeV is emitted, or the de-excitation occurs in two steps, with one photon of energy 1.02 − 0.83 = 0.19 MeV being emitted, followed by a second photon of energy 0.83 MeV.

1




1

Name: ………………………………. Date: ………………………

Radioactive decay - answers 1 260 s or 4 min 20 s

2 0.3 µg

3 a The graph should show activity on the y-axis and time on the x-axis.

b i half-life ≈ 33.5 – 8.5 = 25 mins

ii half-life ≈ 48 – 23 = 25 mins

c 50 mins

4 a 0.0485 days−1

b 34.5 days

5 6.6 × 106 Bq

6 200 litres

7 a 4.6 × 10−4 s−1

b 2.3 × 105

c 25 mins

2




1

Nuclear radius, density and energy levels - answers

Nuclear radius and density

1 4 × 10−14 m

2 2.8 × 10−14 m

3 radius = 2.9 × 10−15 m

density = 2.3 × 1017 kg m−3

4 1.9 × 10−15 m

Nuclear energy levels

5 The excited aluminium nucleus emits gamma photons in order to return to the ground state. Either a single photon of energy 1.02 MeV is emitted, or the de-excitation occurs in two steps, with one photon of energy 1.02 − 0.83 = 0.19 MeV being emitted, followed by a second photon of energy 0.83 MeV.

3




1

Name: ………………………………. Date: ………………………

Radioactive decay - answers 1 260 s or 4 min 20 s

2 0.3 µg

3 a The graph should show activity on the y-axis and time on the x-axis.

b i half-life ≈ 33.5 – 8.5 = 25 mins

ii half-life ≈ 48 – 23 = 25 mins

c 50 mins

4 a 0.0485 days−1

b 34.5 days

5 6.6 × 106 Bq

6 200 litres

7 a 4.6 × 10−4 s−1

b 2.3 × 105

c 25 mins

4

1 © Chris Hamper, InThinking www.physics-inthinking.co.uk

Atomic models 1. Give two pieces of evidence that supports the theory that matter is made of atoms.

2. According to the Thomson model of the Atom (plum pudding/currant bun) what was the expected result of the Rutherford experiment?

3. A beam of electrons are projected into an electric field as shown. Calculate the force on an electron and draw a possible path of the electron.

4. Why can’t electrons orbit the nucleus in circular paths?

Formulae E=V/d F=Eq e = 1.6x10-19 C

5

1

© Chris Hamper, InThinking www.physics-inthinking.co.uk

Exponential Decay

1. The diagram shows an exponential decay curve plotted in LoggerPro.

(a) From the graph estimate the half life of the decay. (b) When there are 37 nuclei the rate of decay = 17 decays/s. Calculate the rate of decay when there are 15 Nuclei. (c) From the information on the graph find the decay constant. (d) Use the decay constant to calculate the half life. 2. A sample of a radioactive isotope decays at a rate of 100 decays per second. If the half life of the material is 2 years how long will it take before the decay rate is 25 decays per second?

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1

© Chris Hamper, InThinking www.physics-inthinking.co.uk

Photoelectric effect 1. Use the quantum model of light to explain why: (a) The KE of photo electrons depends on the frequency not the amplitude.

(b) No photoelectrons are emitted if the frequency is below a certain threshold frequency.

(c) Even in dim light there is no noticeable time delay between switching on the light and the emission of electrons.

2. A photoelectron with KE = 8x10-18 J in emitted from a metal. (a) Calculate the KE of the electron in eV.

(b) What stopping potential would be required to stop the electron?

(c) If the work function of the metal is 10eV calculate the energy of the absorbed photon.

(d) Calculate the frequency of the absorbed photon.

Formulae

E=hf

KEmax = hf – φ

h = 6.6 x 10-34

Js

e=1.6x10-19

C

7

1

E N D - O F -T O P I C Q U E S T I O N S

© Oxford University Press 2014: this may be reproduced for class use solely for the purchaser’s institute

Solutions for Topic 12 – Quantum and nuclear physics (AHL)1. a)

0

lower thresholdfrequency

Emax

0 f

b) electron ejected when energy of light E = hf is greater than the work function (threshold frequency) of the metal

2. a) (i) no change in ammeter

(ii) reading will increase as more electrons are being produced per second

b) (i) photon energy is the sum of the energy of emitted electron plus the work function

(ii) φ = hc _ λ – eV = 6.4 × 10–20 J

3. a) why certain transitions are more likely to occur than others

b) electrons in an atom exist in stationary states; electrons may move from one stationary state to another by absorbing or emitting a quantum of electromagnetic radiation

c) substitute r into equation for En to get required expression

d) binding energy of that level

e) electron described by wavefunction; energy levels can be thought of as standing waves in a potential well; harmonics of standing wave give different levels

4. a) uncertainty in position is ± half of slit width so minimum uncertainty

∆p = 6.63 × 10–34 ___

4π × 0.01 × 10–3 × 0.5 = 1.1 × 10–29 N s

b) component parallel to gap

5. a) particle brought to rest when all kinetic energy has been converted to potential energy, so e.p.e. is 3.8 MeV = 6.1 × 10-13 J

b) Z = 46 so d = kZe × 2e _ Eα

= 3.5 × 10-14 m

c) higher Z so d would be greater

d) (i) mass = A × u

(ii) ρ = m _ v = 3u __ 4π × (1.2 ×10–15)3

= 2 × 1017 kg m-3

6. a) nucleus or atom with speci!c number of protons and neutrons

b) beta minus decay so neutron turns into proton so atomic number must decrease by one; electron antineutrino also produced

c) (i)

time/days8

16

32

64

16 24

activ

ity/1

05 Bq

8

admin

Text Box

2a ii. Reading will decrease as less electrons are being produced. This is due to an increase in frequency requires less photons to be emitted to keep the intensity unchanged.

2

E N D - O F -T O P I C Q U E S T I O N S

© Oxford University Press 2014: this may be reproduced for class use solely for the purchaser’s institute

(ii) λ = ln2 _ t 1 _ 2 = 0.087 day–1

(iii) A = A0 e –λt

t = – 1 _ 0.087

ln 0.5 _ 6.4

= 29 days

7. a) (i) proton or neutron

(ii) proton: uud; neutron: udd

(iii) argon-39 has more neutrons so there is a higher chance of it undergoing beta decay by changing a neutron into a proton

b) (i) Z = 19; N = 39; x is electron antineutrino

(ii) emitted electrons have continuous energy spectrum, so third particle must be produced so that momentum and mass-energy are conserved

(iii) (38.96431 – 38.96370) × u × c2 = 9.1 × 10-14 J

c) (i) separate pure sample of nuclide in a known chemical form, measure its mass then take count rate; use dimensions of G-M tube to calculate activity

(ii) use mass and activity of sample to !nd half life

8. a) separate pure sample of nuclide in a known chemical form, measure its mass then take count rate; use dimensions of G–M tube to calculate activity

b) fraction remaining = 2–1.6 = 0.33

9

DBT�MPOH�BT��

10

2 PHYSICS FOR THE IB DIPLOMA © CAMBRIDGE UNIVERSITY PRESS 2015ANSWERS TO EXAM-STYLE QUESTIONS – TOPIC 12

c i Extending the graph to the vertical intercept gives –3.4 V. ! So the work function is 3.4 eV. !

ii From E hf= − φ and E eV= we have that Vhef

e= − φ and so the gradient of the graph is the Planck

constant divided by e. !

The gradient is 8.0 03.0 10 0.90 10

3.8 10 VHz15 1515 1−

× − ×= × − − . !

And so h 1.6 10 3.8 10 6.1 10 CVHz 6.1 10 J s19 15 34 1 34= × × × = × = ×− − − − − . !

iii The threshold frequency is 0.90 10 Hz15× . !

And so the maximum wavelength is 3.0 100.90 10

3.3 10 m8

157×

×= × − . !

d The energy of the emitted electrons does not depend on intensity. ! So the graph will not change. !

12 a The net force on the electron is the electric force of attraction between the electron and the proton i.e. ker

2

2 . !

Equating this with the centripetal force mvr

2 gives the answer. !

b The Bohr condition is that mvr nh

2π= . !

Squaring gives m v r nh4

2 2 2 22

2π= and substituting the expression from the previous part leads to

mkemr

r nh4

22

2 22

2π= . !

Simplifying gives mke r nh4

2 22

2π= and the answer. !

c The total energy of the electron is E mvker

12

22

= − . !

Substituting the value for the square of the speed in the "rst part again gives the answer. ! d It signi"es that the electron is bound to the proton and cannot escape far away unless su#cient energy is

provided to it. !

e From hp

λ = we "nd phλ

= and so the Bohr condition becomes hr n

h2λ π

= . !

Simplifying gives the answer. ! f i An electron wave is a wave whose amplitude is related to the probability of "nding the electron somewhere

in space at a given time. ! ii The wave corresponds to n = 4. !

From b, r nhmke4

4(6.63 10 )

4 9.11 10 8.99 10 (1.6 10 )2

2

2 2

34 2

2 31 9 19 2π π= = × ×

× × × × × ×

−

− − . !

r 2.1 10 m10= × − . !

iii The total energy from c is Eker

= − = × × ×× ×

=−

−

2 9 19 2

1028 99 10 1 6 10

2 2 1 105

. ( . ).

..5 10 19× − J and this, or more, is what must be supplied. !

g The probability wave associated with the electron implies that the electron is not an object that is localised at a particular point at a given time, !

but can be thought to be spread out through space like waves do. ! The Bohr orbit only gives the average position of the electron. !

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ANSWERS TO EXAM-STYLE QUESTIONS – TOPIC 12 3PHYSICS FOR THE IB DIPLOMA © CAMBRIDGE UNIVERSITY PRESS 2015

13 a To every particle there corresponds a wave of probability. ! With a wavelength that is given by the Planck constant divided by the momentum of the particle. !

b i qV Epm2K

2

= = !

Hence p mqV2= and the result follows from hp

λ = . !

ii hmqV2

6.63 10

2 9.11 10 1.6 10 1201.1 10 m

34

31 1910λ = = ×

× × × × ×= ×

−

− −− !

c In a Davisson-Germer type of experiment electrons that have been accelerated are directed at a crystal from which they di$ract and interfere. !

From the interference pattern the wavelength may be determined. ! And this is consistent with the de Broglie formula. !

d The de Broglie wavelength of the bullet is hp

6.63 100.080 420

2 10 m34

35λ = = ××

≈ ×−

− . !

For di$raction e$ects to be seen the wavelength must be comparable to the size of the hole. ! But 2 10 m 5.0 cm35× <<− . ! And so no di$raction will be observed. !

14 a Tunnelling is a quantum mechanical phenomenon in which particles can be transmitted through energy barriers. !

That would classically be impossible due to energy conservation. ! b i The width is about 2.8 10 1.3 10 1.5 10 m10 10 10× − × = ×− − − . ! ii From the graph the de Broglie wavelength before and after is the same. ! And hence the ratio is 1. ! iii The wavefunction squared is proportional to the probability of "nding a particle somewhere. !

And so the transmitted ratio is 320

2 102

2

≈ × − . !

c Protons have a higher mass so fewer of them would get transmitted. !

15 a i The electron antineutrino. ! ii Electrically neutral. ! Very small non-zero mass. ! b If no third particle were present in the products of the beta decay the electron would always carry away a "xed

proportion of the total energy released. ! But experiments show that this is not the case which means a third particle must be sharing in the energy. ! c When the tree dies it will no longer absorb C-14 from its surroundings. ! The amount of C-14 present when the tree died will then diminish with time because C-14 is unstable and

decays into N-14. ! d We may ignore C-14 in this part of the calculation since its concentration is so small. !

So 15 g correspond to 1512

6.02 10 7.525 10 7.5 1023 23 23× × = × ≈ × atoms. !

e A N14λ= × and so NA

14 λ= with

ln 25730 365 24 3600

3.8359 10 s12 1λ =× × ×

= × − − . !

N

A 1.403.8359 10

3.6498 10 3.6 1014 1211 11

λ= =

×= × ≈ ×− . !

Hence NN

3.6498 107.525 10

4.85 1014

12

11

2313= ×

×= × − . !

12

4 PHYSICS FOR THE IB DIPLOMA © CAMBRIDGE UNIVERSITY PRESS 2015ANSWERS TO EXAM-STYLE QUESTIONS – TOPIC 12

f NN

e4.85 10 1.3 10 t14

12

13 12= × = × λ− − − so e t− =λ 0 3731. !

t tln0.3731

ln0.3731λλ

− = ⇒ = − !

t = −

×= × ≈−

ln ..

.0 3731

3 8359 102 57 10 815012

11 s year !

16 a Alpha particle and gamma ray energies in radioactive decay, ! are discrete. ! b i Correct transition selected. !

energy/MeV

0.294 0.1420.0430.0

5.902

ii 5.902 – 0.043 = 5.86 MeV ! c i The nuclear force has a short range. ! And is practically zero for distances larger than the nuclear radii. ! ii It must overcome an energy barrier of height 30 MeV and its total energy is less than this. ! Leaving the nucleus would violate energy conservation. ! iii Like all particles alpha particles have wavelike properties and are described by quantum mechanical

wavefunctions. ! Which allow for the tunneling phenomenon in which the wavefunction leaks out into the classically

forbidden region. ! d The half-life has to do with the tunneling probability, i.e. how long an alpha particle takes to leave the nucleus

on the average. ! And this tunneling probability is very sensitive to small changes in alpha particle energies. ! e The uncertainty in position is of order x 10 m15∆ ≈ − . !

Hence the uncertainty in momentum is phx

h4 4 10

5.3 10 N s1520

π π∆ ≈

∆≈

×= ×−

− . !

17 a i b

bsinsin

θ λ λθ

= ⇒ = !

E

pm

p E m2

2 2 54 10 1.6 10 1.67 10 1.7 10 N sK

2

K6 19 27 19= ⇒ = = × × × × × × = ×− − − !

Hence 6.63 101.7 10

3.9 10 m34

1915λ = ×

×= ×

−

−− and then b = ×

°= ×

−−3 9 10

151 5 10

1514.

sin. m. !

ii m A 1.661 10 kg27≈ × × − . !

V A A

43

1.2 10 7.24 10 m15 3 45 313

π )(= × × = × ×− − . !

mV

AA

1.661 107.24 10

2.29 10 2 10 kgm27

4517 17 3ρ = = × ×

× ×= × ≈ ×

−

−− . !

b EkQqd

= !

d

8.99 10 2 82 1.6 105.2 10

9 19

6= × × × × ××

−. !

d 4.54 10 4.5 10 m14 14= × ≈ ×− − !

c i The only force acting on the alpha particle is the electric force. ! ii A sharp decrease in the number of scattered particles at high energies. ! As the energy increases the alpha particles approach closer to the nucleus and so the nuclear force acts on

them, the nucleus absorbs some thus reducing the number that is being scattered. !

13

Topic 12 (New) [113 marks]

1a.

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

antineutrino AND charge AND mass number of electron ,

conservation of mass number AND charge ,

Do not accept V.

Accept without subscript e.

[2 marks]

104 Be →10

5 B + 0−1 e + ¯¯Ve

0−1e ¯¯V

105 B 10

4 Be

V

1b.

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.

0

[2 marks]

[2 marks]

14

Markschemecorrect shape ie increasing from 0 to about 0.80 N

crosses given line at 0.50 N

[2 marks]

0

0

1c. 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]

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

1d. Beryllium-10 is used to investigate ice samples from Antarctica. A sample of ice initiallycontains 7.6 × 10 atoms of beryllium-10. The present activity of the sample is 8.0 ×10 Bq.

Determine, in years, the age of the sample.

11

−3

[3 marks]

[3 marks]

15

Markschemeλ «=

» = 4.95 × 10 «y »

rearranging of A = λN e to give –λt = ln «= –0.400»

t = «y»

Allow ECF from MP1

[3 marks]

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

1e.

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]

1f. 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]

1g. Calculate the peak wavelength in the intensity of the radiation emitted by the icesample.

[1 mark]

[2 marks]

[2 marks]

16

Markscheme

= 1.1 × 10 «m»

Allow ECF from MP1 (incorrect temperature).

[2 marks]

λ = 2.90×10−3

253

–5

1h. 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]

2a.

Hydrogen atoms in an ultraviolet (UV) lamp make transitions from the first excited state to theground state. Photons are emitted and are incident on a photoelectric surface as shown.

Show that the energy of photons from the UV lamp is about 10 eV.

[2 marks]

[2 marks]

17

MarkschemeE = –13.6 «eV» E = – = –3.4 «eV»

energy of photon is difference E – E = 10.2 «≈ 10 eV»

Must see at least 10.2 eV.

[2 marks]

1 213.6

4

2 1

2b.

The photons cause the emission of electrons from the photoelectric surface. The work functionof the photoelectric surface is 5.1 eV.

Calculate, in J, the maximum kinetic energy of the emitted electrons.

Markscheme10 – 5.1 = 4.9 «eV»

4.9 × 1.6 × 10 = 7.8 × 10 «J»

Allow 5.1 if 10.2 is used to give 8.2×10 «J».

–19 –19

−19

2c. Suggest, with reference to conservation of energy, how the variable voltage sourcecan be used to stop all emitted electrons from reaching the collecting plate.

MarkschemeEPE produced by battery

exceeds maximum KE of electrons / electrons don’t have enough KE

For first mark, accept explanation in terms of electric potential energy difference ofelectrons between surface and plate.

[2 marks]

2d. The variable voltage can be adjusted so that no electrons reach the collecting plate.Write down the minimum value of the voltage for which no electrons reach the collectingplate.

[2 marks]

[2 marks]

[1 mark]

18

Markscheme4.9 «V»

Allow 5.1 if 10.2 is used in (b)(i).

Ignore sign on answer.

[1 mark]

2e.

The electric potential of the photoelectric surface is 0 V. The variable voltage is adjusted so thatthe collecting plate is at –1.2 V.

On the diagram, draw and label the equipotential lines at –0.4 V and –0.8 V.

Markschemetwo equally spaced vertical lines (judge by eye) at approximately 1/3 and 2/3

labelled correctly

[2 marks]

2f. An electron is emitted from the photoelectric surface with kinetic energy 2.1 eV.Calculate the speed of the electron at the collecting plate.

[2 marks]

[2 marks]

19

Markschemekinetic energy at collecting plate = 0.9 «eV»

speed = «

» = 5.6 × 10 «ms »

Allow ECF from MP1

[2 marks]

√ 2×0.9×1.6×10−19

9.11×10−315 –1

3a. 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]

h2π

h2π

3b. 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.

Markscheme

OR

KE = PE hence m v =

«solving for v to get answer»

Answer given – look for correct working

[1 mark]

v = √ ke2

mer

=mev2

rke2

r2

12

12 e

2 12

ke2

r

[3 marks]

[1 mark]

20

3c. 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.

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]

r =h2

4π2kmee2

√ ke2

mer eh

2π

me2 r2 =ke2

merh2

4π2

3d. Calculate the electron’s orbital radius in (c)(ii).

Markscheme« r =

»

r = 5.3 × 10 «m»

[1 mark]

(6.63×10−34)2

4π2×8.99×109×9.11×10−31×(1.6×10−19)2

–11

3e.

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.

10645Rh 106

46Pd –

–

–

[2 marks]

[1 mark]

[3 marks]

21

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]

3f. Suggest why the β decay is followed by the emission of a gamma ray photon.

Markschemethe palladium nucleus emits the photon when it decays into the ground state «from theexcited state»

[1 mark]

–

3g. 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]

6 –19 –14

=hcE

6.63×10−34×3×108

7.68×10−14–12

4a.

The Feynman diagram shows electron capture.

State and explain the nature of the particle labelled X.

[1 mark]

[2 marks]

[3 marks]

22

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

4b.

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

4c. 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

∝ 1√E

∝ 1E

[2 marks]

[3 marks]

23

4d. 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

4e.

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.

Markschemecorrectly plotted

Allow ECF from (d)(i)

–15

4f. Draw a line on the graph, to show the variation of nuclear radius with nucleon number.

[2 marks]

[1 mark]

[2 marks]

24

Markschemesingle smooth curve passing through both points with decreasing gradient

through origin

5a.

Two observations about the photoelectric effect are

Observation 1: For light below the threshold frequency no electrons are emitted from the metalsurface.

Observation 2: For light above the threshold frequency, the emission of electrons is almostinstantaneous.

Explain how each observation provides support for the particle theory but not the wavetheory of light.

[4 marks]

25

MarkschemeObservation 1:particle – photon energy is below the work functionORE = hf and energy is too small «to emit electrons»wave – the energy of an em wave is independent of frequency

Observation 2:particle – a single electron absorbs the energy of a single photon «in an almostinstantaneous interaction»wave – it would take time for the energy to build up to eject the electron

5b.

The graph shows how the maximum kinetic energy E of electrons emitted from a surface ofbarium metal varies with the frequency f of the incident radiation.

Determine a value for Planck’s constant.

max

[2 marks]

26

Markschemeattempt to calculate gradient of graph = « »

«Js»

Do not allow a bald answer of 6.63 x 10 Js or 6.6 x 10 Js.

4.2×10−19

6.2×1014

= 6.8 − 6.9 × 10−34

-34 -34

5c. State what is meant by the work function of a metal.

MarkschemeALTERNATIVE 1minimum energy required to remove an electron «from the metal surface»

ALTERNATIVE 2energy required to remove the least tightly bound electron «from the metal surface»

5d. Calculate the work function of barium in eV.

MarkschemeALTERNATIVE 1reading of y intercept from graph in range 3.8 − 4.2 × 10 «J»conversion to eV = 2.4 – 2.6 «eV»

ALTERNATIVE 2reading of x intercept from graph «5.8 − 6.0 × 10 Hz» and using hf to get 3.8 − 4.2 × 10

«J»conversion to eV = 2.4 – 2.6 «eV»

–19

14 0

–

19

5e. The experiment is repeated with a metal surface of cadmium, which has a greaterwork function. Draw a second line on the graph to represent the results of thisexperiment.

Markschemeline parallel to existing lineto the right of the existing line

The first scientists to identify alpha particles by a direct method were Rutherford and Royds.

[1 mark]

[2 marks]

[2 marks]

27

6a.

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.

22686 Ra

42α

42He22286 Rn

6b.

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]

[2 marks]

28


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 .

5

A0e−1.4×10−11×5.8×105≈ A0

0 λ 0

A

A0

0–15 –1

11 10

λ –1

6c. Show that about 3 x 10 alpha particles are emitted by the radium-226 in 6 days.15 [3 marks]

29


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

λ 0

A20

9 –1

–11 20 15

0e−λt

0 A20

0 0(1 − e−λ×6×24×3600) 15

6d.

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.

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.

6e. 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.

–5 3

[1 mark]

[3 marks]

30

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).

–9 2911.3×10−5

15 –23 2911.3×10−5

15

7a.

Yellow light of photon energy 3.5 x 10 J is incident on the surface of a particular photocell.

Calculate the wavelength of the light.

Markschemewavelength = « » 5.7 x 10 «m»

If no unit assume m.

–19

= =hcE

1.99×10−25

3.5×10−19–7

7b. Electrons emitted from the surface of the photocell have almost no kineticenergy. Explain why this does not contradict the law of conservation of energy.

Markscheme«potential» energy is required to leave surface

Do not allow reference to “binding energy”.Ignore statements of conservation of energy.

all/most energy given to potential «so none left for kinetic energy»

7c. Radiation of photon energy 5.2 x 10 J is now incident on the photocell. Calculatethe maximum velocity of the emitted electrons.

–19

[1 mark]

[2 marks]

[2 marks]

31

Markschemeenergy surplus = 1.7 x 10 J

v = «m s »

Award [1 max] if surplus of 5.2 x 10 J used (answer: 1.1 x 10 m s )

–19

max √ = 6.1 × 1052×1.7×10−19

9.1×10−31–1

–19 6 –1

7d.

The photocell is connected to a cell as shown. The photoelectric current is at its maximum value(the saturation current).

Radiation with a greater photon energy than that in (b) is now incident on the photocell. Theintensity of this radiation is the same as that in (b).

Describe the change in the number of photons per second incident on the surface of thephotocell.

Markscheme«same intensity of radiation so same total energy delivered per square metre per second»

light has higher photon energy so fewer photons incident per second

Reason is required

7e. State and explain the effect on the maximum photoelectric current as a resultof increasing the photon energy in this way.

[1 mark]

[3 marks]

32

Markscheme1:1 correspondence between photon and electron

so fewer electrons per second

current smaller

Allow ECF from (c)(i)Allow ECF from MP2 to MP3.

8a.

An apparatus is used to investigate the photoelectric effect. A caesium cathode C is illuminatedby a variable light source. A variable power supply is connected between C and the collectinganode A. The photoelectric current I is measured using an ammeter.

A current is observed on the ammeter when violet light illuminates C. With V heldconstant the current becomes zero when the violet light is replaced by red light of thesame intensity. Explain this observation.

Markschemereference to photonORenergy = hf or =

violet photons have greater energy than red photons

when hf > Φ or photon energy> work function then electrons are ejected

frequency of red light < threshold frequency «so no emission»ORenergy of red light/photon < work function «so no emission»

hcλ

The graph shows the variation of photoelectric current I with potential difference V

[3 marks]

33

8b. The graph shows the variation of photoelectric current I with potential difference Vbetween C and A when violet light of a particular intensity is used.

The intensity of the light source is increased without changing its wavelength.

(i) Draw, on the axes, a graph to show the variation of I with V for the increased intensity.

(ii) The wavelength of the violet light is 400 nm. Determine, in eV, the work function of caesium.

(iii) V is adjusted to +2.50V. Calculate the maximum kinetic energy of the photoelectrons justbefore they reach A.

Markschemeiline with same negative intercept «–1.15V»

otherwise above existing line everywhere and of similar shape with clear plateau

Award this marking point even if intercept is wrong.

ii

« » 3.11 «eV»

Intermediate answer is 4.97×10 J.

Accept approach using f rather than c/λ

«3.10 − 1.15 =» 1.96 «eV»Award [2] for a bald correct answer in eV.Award [1 max] if correct answer is given in J (3.12×10 J).

iii

«KE = qVs =» 1.15 «eV»

OR

1.84 x 10 «J»

Allow ECF from MP1 to MP2.

adds 2.50 eV = 3.65 eV

OR

5.84 x 10 J

Must see units in this question to identify energy unit used.Award [2] for a bald correct answer that includes units.Award [1 max] for correct answer without units.

=hcλe

=6.63×10−34×3×108

40×10−9×1.6×10−19

−19

−19

−19

−19

[6 marks]

34

9a. A particular K meson has a quark structure s. State the charge, strangeness andbaryon number for this meson.

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.

u

−

9b. 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.

–

[2 marks]

[3 marks]

35

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.

– –

+ + + –

9c. 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.

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.

0–λt

n

0.21 = e−( )ln2×t

5700

An alpha particle with initial kinetic energy 32 MeV is directed head-on at a nucleus of

[4 marks]

36

10a. An alpha particle with initial kinetic energy 32 MeV is directed head-on at a nucleus ofgold-197 .

(i) Show that the distance of closest approach of the alpha particle from the centre of thenucleus is about 7×10 m.

(ii) Estimate the density of a nucleus of using the answer to (a)(i) as an estimate of thenuclear radius.

Markscheme(i)32 MeV converted using 32×10 ×1.6×10 «=5.12×10 J»

OR 7.102×10 m

«d≈7×10 m»

Must see final answer to 2+ SF unless substitution is completely correct with value for kexplicit.Do not allow an approach via .

(ii)m≈197×1.661×10OR3.27×10 kg

OR

1.44×10 m

Allow working in MeV: 1.28×10 MeVc m .Allow ECF from incorrect answers to MP1 or MP2.

(19779 Au)

−15

19779 Au

6 –19 –12

d =≪ = =≫kQq

E

8.99×109×2×79×(1.6×10−19)2

32×106×1.6×10−198.99×109×2×79×1.6×10−19

32×106

-15

-15

r = R0A13

-27

-25

V = × (7 × 10−15)34π

3

-42 -3

ρ =≪ = =≫ 2.28 × 1017 ≈ 2 × 1017kgm−3mV

3.2722×10−25

1.4368×10−42

47 –2 –3

10b. The nucleus of is replaced by a nucleus of the isotope . Suggest thechange, if any, to your answers to (a)(i) and (a)(ii).

Distance of closest approach:

Estimate of nuclear density:

MarkschemeDistance of closest approach: charge or number of protons or force of repulsion is thesame so distance is the same

Estimate of nuclear density: « so» density the same

19779 Au 195

79 Au

ρ ∝ A

(A )313

10c. An alpha particle is confined within a nucleus of gold. Using the uncertainty principle,estimate the kinetic energy, in MeV, of the alpha particle.

[5 marks]

[2 marks]

[3 marks]

37

Printed for Concordian International School

© International Baccalaureate Organization 2019 International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®

MarkschemeΔx≈7×10 m

Accept Δx≈3.5×10 m or Δx≈1.4×10 m leading to E≈0.11MeV or 0.0067MeV.Answer must be in MeV.

–15

Δp ≈ ≪= 7.54 × 10−21Ns ≫6.63×10−34

4π×7×10−15

E ≈≪ = = 4.3 × 10−15J = 26897eV ≫≈ 0.027MeVΔp2

2m

(7.54×10−21)2

2×6.6×10−27

–15 –14

38

Markscheme-Topic 12: Quantum and nuclear physics

Markscheme-Topic 12.1: The interaction of matter with radiation

1. B [1]

2. C [1]

3. B [1]

4. C [1]

5. B [1]

6. A [1]

7. D [1]

8. A [1]

9. C [1]

10. B [1]

11. B [1]

12. B [1]

13. (a) photoelectric current / rate of emission independent of frequency; photoelectric current / rate of emission depends on intensity of radiation; (max) kinetic energy of electron dependent on frequency; existence of threshold frequency; instantaneous ejection; etc; 3 max

(b) (i) hf = hf0 + eVs;

Accept ϕ instead of hf0.

identifies h and e; identifies f0 / ϕ; 3

(ii) re-arranging, VS = × f – × f0;

compares with y = mx +c and hence gradient ; 2

(iii) f0 = 0.96 × 1015 Hz; work function = 6.6 × 10–34 × 0.96 × 1015 = 6.3 × 10–19 J / 3.9 eV; 2

[10]

14. (a) all particles have a wavelength associated with them / OWTTE;

the de Broglie hypothesis gives the associated wavelength as λ = ;

where h is the Planck constant and p is the momentum of the particle; 3

If answers just quote the formula from the data book then

eh

eh

eh

ph

39


award [1] for showing at least they recognize which formula relates to the hypothesis.

(b) (i) KE = Ve = 850 × 1.6 × 10–19 J = 1.4 × 10–16 J; 1

(ii) use E = to get p = ;

substitute p = = 1.6 × 10–23 N s; 2

(iii) λ = ;

substitute λ = = 4.1 × 10–11 m; 2

[8]

15. (a) fine structure / relative intensity of the spectral lines; 1

(b) an electron only changes orbit if it emits or absorbs a photon / energy; the energy (of the emitted or absorbed photon) is equal to the difference in the energies of the (permitted) orbits / OWTTE; 2

(c) substitute for r into En = – to get En = – ;

therefore, En = – ;

where K = – ; 3

(d) energy; and is the amount required to remove an electron from the orbit n = 1 to n = ∞; or the total energy of the electron; when n = 1 / in the ground state; 2

(e) the electrons are described by wave functions; the wave functions can only have certain values because they have to fit boundary conditions / the wave functions behave like standing waves so can only have certain values / OWTTE; 2

Be generous – award [1] for the idea of wave function and [1] for some other related physics.

[10]

16. (a) use of 4hx pπ

∆ ⋅∆ =

to give ∆p = 0.6(or 1.1) × 10−29Ns; 2 accept ∆x being one or half slit width

(b) normal to direction of beam / parallel to plane of slit; 1 [3]

17. (a) a function whose (absolute squared) value may be used to calculate the probability of finding a particle near a given position / quantity related to the probability of finding an electron near a given position/at a given position; 1

(b) middle of the box / (near) 0.5 × 10–10 m; 1

(c) the de Broglie wavelength is 2.0 × 10–10 m;

mp2

2

mE2

1631 104.1101.92 −− ××××

ph

23

34

106.1106.6

−

−

××

rke2

2

22

422mπ2hn

ek

2nK

2

422mπ2h

ek

40


p = = 3.3 × 10–24 Ns; 2

(d) difference in energy is

∆E = 1.635 × 10–18 J;

λ = ;

λ = = 1.22 × 10–7 m; 3

(e) (i) attempt at using the energy – time uncertainty relation;

∆E = 5.3 × 10–25 J; 2

(ii) the wavelength of the photons is determined by the difference in energy between the two levels; and that energy difference is not well defined/definite/not always the same (because of the uncertainty principle); 2

(f) energy levels all with strictly positive energy; difference between levels increasing with increasing n; 2

Judge separation of levels by eye – there will not be numbers on the candidates’ graphs.

[13]

10

34

102.21063.6

−

−

××

=λh

×+

×−=

−−

2

18

2

18

11018.2

21018.2

Ehc∆

××××

−

−

18

834

10635.1100.31063.6

×××

=∆

=−

−

10

34

100.141063.6

π4 πth

41


Markscheme-Topic 12.2: Nuclear physics

1. B [1]

2. C [1]

3. B [1]

4. A [1]

5. B [1]

6. B [1]

7. B [1]

8. C [1]

9. A [1]

10. C [1]

11. B [1]

12. B [1]

13. C [1]

14. D [1]

15. (a) the kinetic energy of 3.8 MeV gets converted to electrical potential energy;

equal therefore to 3.8 × 106 × 1.6 × 10−19 = 6.1 × 10−13J; 2

(b) correct identification of charges of alpha and palladium nucleus;

electrical potential energy =

d = 3.5 × 10−14 m; 3

Award [1 max] for those who fail to square the elementary charge (answer 2.2 × 103 m) or those who square the d in the denominator (answer 1.9 × 10−7 m).

(c) it will be greater since the gold nucleus has more protons / force of repulsion is greater / electrical potential energy between alpha and gold nucleus is larger at the same distance; 1

( ) ;106.1462109219

9

d

−×××××

42


(d) (i) 1

(ii)

=

to give kg m−3 (no mark for answer) 1 [8]

16. (a) (i) time for the activity to halve in value / time for the number of nuclei to transmute to nuclei of another element / OWTTE; 1

(ii) the probability that a nucleus will decay in unit time; 1

(b) use of N = Noe–λt with N = No;

to give from which = ln 2; 2

(c) λ = = 0.039 d–1;

substitute into A = Aoe–λt to get A = 4.5 × 104 Bq; 2

(d) (i) mass defect = 227.0278 – (223.0186 + 4.0026) =0.0066 u; = 6.148 MeV c–2; therefore energy of γ = 6.148 – 5.481 = 0.667 MeV; 3

(ii) use f = ;

E = 0.667 × 106 × 1.60 × 10–19 J; to give f = 1.62 × 1020 Hz; 3

(e)

(i) two correct As; 1

(ii) two correct Gs; 1

(iii) correct R; 1

(f) 5.481 + 0.667 = 6.148 MeV; 1 [16]

;Aum≈

Vm

=ρ

( );

m102.13

4kg1066.1

3315

27

−

−

×π

××

A

A

17102×≈ρ

21

21=2

1T

eλ

21Tλ

182ln

hE

43


17. (a) (i) (electron) anti-neutrino; 1

(ii) lepton; 1

(b) (i) 1

(ii) 0.075 = 0.24 × ;

t = 9.7 × 103 year; 3

(c) measure activity of source; determine number of molecules chemically; activity = λ × N , hence half-life; 3

Award [1 max] for method that measures activity and then waits before re-measuring.

[9]

18.

0

34 815

6 19

1515

0

=356.6 10 3.0 10= 3.1 10400 10 1.6 10

3.1 10 5.4 10sin sin 35

hc mE

D m

θ

λ

λθ

−−

−

−−

× × ×= = ×

× × ××

= = = ×

The radius is 2.7×10-15m. 2 [2]

;5700

2ln=λ

t××− −41021.1e

;101.2175

240ln

4−×

=t

44

Documents

12 Quantum and nuclear physics