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Question 21M.2.HL.TZ1.7

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Date May 2021 Marks available [Maximum mark: 11] Reference code 21M.2.HL.TZ1.7
Level HL Paper 2 Time zone TZ1
Command term Compare and contrast, Determine, Identify, Outline, Write down Question number 7 Adapted from N/A
7.
[Maximum mark: 11]
21M.2.HL.TZ1.7

Radioactive uranium-238 U92238 produces a series of decays ending with a stable nuclide of lead. The nuclides in the series decay by either alpha (α) or beta-minus (β) processes.

(a)

Uranium-238 decays into a nuclide of thorium-234 (Th).


Write down the complete equation for this radioactive decay.

[1]

Markscheme

U→«92238»90234Th+«»24α ✓ 

Allow He for alpha.

(b)

Thallium-206 Tl81206 decays into lead-206 Pb82206.

Identify the quark changes for this decay.

[1]

Markscheme

udd→uud
OR
down quark changes to up quark  

(c)

The half-life of uranium-238 is about 4.5 × 109 years. The half-life of thallium-206 is about 4.2 minutes.

Compare and contrast the methods to measure these half-lives.

[4]

Markscheme

measure «radio»activity/«radioactive» decay/A for either
OR
take measurements with a Geiger counter.

for Uranium measure number/N of radioactive atoms/OWTTE

for Thalium measure «rate of» change in activity over time.

correct connection for either Uranium or Thalium to determine half life ✓ 

The graph shows the variation with the nucleon number A of the binding energy per nucleon.

(d.i)

Outline why high temperatures are required for fusion to occur.

 

[2]

Markscheme

links temperature to kinetic energy/speed of particles  

energy required to overcome «Coulomb» electrostatic repulsion  

(d.ii)

Outline, with reference to the graph, why energy is released both in fusion and in fission.

 

[1]

Markscheme

«energy is released when» binding energy per nucleon increases

(d.iii)

Uranium-235 U92235 is used as a nuclear fuel. The fission of uranium-235 can produce krypton-89 and barium-144.

Determine, in MeV and using the graph, the energy released by this fission.

[2]

Markscheme

any use of (value from graph) x (number of nucleons)

«235 × 7.6 – (89 × 8.6 + 144 × 8.2) =» 160 «MeV»