DP Physics (first assessment 2025)

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Question 20N.2.HL.TZ0.10

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Date November 2020 Marks available [Maximum mark: 9] Reference code 20N.2.HL.TZ0.10
Level HL Paper 2 Time zone TZ0
Command term Discuss, Estimate, Outline, Show that Question number 10 Adapted from N/A
10.
[Maximum mark: 9]
20N.2.HL.TZ0.10

The de Broglie wavelength λ of a particle accelerated close to the speed of light is approximately

λ≈hcE

where E is the energy of the particle.
A beam of electrons of energy 4.2×108 eV is produced in an accelerator.

The electron beam is used to study the nuclear radius of carbon-12. The beam is directed from the left at a thin sample of carbon-12. A detector is placed at an angle θ relative to the direction of the incident beam.

The graph shows the variation of the intensity of electrons with θ. There is a minimum of intensity for θ=θ0.

(a)

Show that the wavelength of an electron in the beam is about 3×10-15 m.

[1]

Markscheme

λ=6.63×10-34×3×1081.60×10-19×4.2×108 OR =2.96×10-15«m» ✓


Answer to at least 2 s.f. (i.e. 3.0)

Examiners report

An easy calculation with only one energy conversion to consider and a 'show' answer to help. 

(b(i))

Discuss how the results of the experiment provide evidence for matter waves.

[2]

Markscheme

«the shape of the graph suggests that» electrons undergo diffraction «with carbon nuclei» ✓

only waves diffract ✓

Examiners report

This question was challenging for candidates many of whom seemed to have little idea of the experiment. Many answers discussed deflection, with the idea that forces between the electron and the nucleus causing it to deflect at a particular angle. This was often combined with the word interference to suggest evidence of matter waves. A number of answers described a demonstration the candidates remembered seeing so answers talked about fuzzy green rings.

(b(ii))

The accepted value of the diameter of the carbon-12 nucleus is 4.94×10-15 m. Estimate the angle θ0 at which the minimum of the intensity is formed.

[2]

Markscheme

sinθ0=2.96×10-154.94×10-15«=0.599» ✓

37 «degrees» OR 0.64/0.65 «rad» ✓

Examiners report

This was answered reasonably well with only the odd omission of the sine in the equation.

(b(iii))

Outline why electrons with energy of approximately 107 eV would be unsuitable for the investigation of nuclear radii.

[2]

Markscheme

the de Broglie wavelength of electrons is «much» longer than the size of a nucleus ✓


hence electrons would not undergo diffraction
OR
no diffraction pattern would be observed ✓

Examiners report

Candidates generally scored poorly on this question. There was confusion between this experiment and another diffraction one, so often the new wavelength was compared to the spacing between atoms. Also, in line with answers to b(i) there were suggestions that the electrons did not have sufficient energy to reach the nucleus or would be deflected by too great an angle to be seen.

(c)

Experiments with many nuclides suggest that the radius of a nucleus is proportional to A13, where A is the number of nucleons in the nucleus. Show that the density of a nucleus remains approximately the same for all nuclei.

[2]

Markscheme

volume of a nucleus proportional to A133=A AND mass proportional to A ✓

the ratio massvolume independent of A «hence density the same for all nuclei» ✓

 

Both needed for MP1

Examiners report

This question proved challenging and it wasn't common to find answers that scored both marks. Of those that had the right approach some missed out on both marks by describing A as the mass of the nucleus rather than proportional to the mass of the nucleus.

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