IB Physics HL E Nuclear and Quantum Physics Questions

Practise IB Physics HL nuclear and quantum physics through shared-core and HL decay, photons, binding energy and particle evidence using precise equations.

Exam points

Combine atomic, isotope, spectral, quantum and nuclear-binding models to interpret detailed HL evidence.
Apply quantum, radioactive-decay and detector relationships with uncertainty, background corrections and model limits.
Calculate fission and fusion energy changes, model chain reactions and evaluate reactor or stellar evidence.
Use luminosity, spectra, Wien, Stefan–Boltzmann, stellar-evolution and parallax relationships to infer stellar properties and distances.

Covered syllabus points

E.1 Structure of the atom E.2 Quantum physics E.3 Radioactive decay E.4 Fission E.5 Fusion and stars

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Study E. Nuclear and quantum physics conceptsOpen the corresponding Concept lesson for explanations, examples and exam connections.

[Maximum number: 12]

(a)

Rutherford constructed a model of the atom based on the results of the alpha particle scattering experiment. Describe this model.

[ 2 ]

(b)

Bohr modified the Rutherford model by introducing the condition $m v r=n \frac{h}{2 \pi}$ .

Outline the reason for this modification.

[ 3 ]

the 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 $mvr=n\frac{h}{2\pi}$

(c)

Calculate the electron's orbital radius in (c)(ii).

[ 1 ]

$r=\frac{(6.63\times10^{-34})^2}{4\pi^2\times8.99\times10^9\times9.11\times10^{-31}\times(1.6\times10^{-19})^2} r=5.3\times10^{-11}\ \mathrm{m}$

(d)

Rhodium-106 $\left({ }_{45}^{106} \mathrm{Rh}\right)$ decays into palladium-106 $\left({ }_{46}^{106} \mathrm{Pd}\right)$ by beta minus $\left(\beta^{-}\right)$ decay. The diagram shows some of the nuclear energy levels of rhodium-106 and palladium-106. The arrow represents the $\beta^{-}$ decay.

[ 6 ]

(i)

Explain what may be deduced about the energy of the electron in the $\beta^{-}$ decay.

[ 3 ]

the energy released is $3.54-0.48=3.06\ \mathrm{MeV}$

this is shared by the electron and the antineutrino

so the electron’s energy varies from 0 to $3.06\ \mathrm{MeV}$

(ii)

Suggest why the $\beta^{-}$ decay is followed by the emission of a gamma ray photon.

[ 1 ]

(iii)

Calculate the wavelength of the gamma ray photon in (d)(ii).

[ 2 ]

Photon energy

$\begin{aligned} E&=0.48\times10^6\times1.6\times10^{-19}=7.68\times10^{-14}\ \mathrm{J}\\ \lambda&=\frac{hc}{E}=\frac{6.63\times10^{-34}\times3\times10^8}{7.68\times10^{-14}}=2.6\times10^{-12}\ \mathrm{m} \end{aligned}$

[Maximum number: 10]

This question is in two parts. Part 1 is about photoelectricity. Part 2 is about electrical and magnetic force fields.
Part 1 Photoelectricity

(a)

State what is meant by work function.

[ 1 ]

(b)

The diagram shows part of an experimental arrangement used to investigate the photoelectric effect.

[ 4 ]

(i)

Explain how the maximum kinetic energy of the emitted electrons is determined experimentally.

[ 2 ]

(ii)

On the diagram, draw the power supply and other necessary components needed in order to carry out the experiment in (b)(i).

[ 2 ]

(c)

Using results obtained with the apparatus in (b), the following graph was drawn. The graph shows how the maximum kinetic energy of the photoelectrons varies with the frequency of the incident radiation.

State how the graph can be used to determine

[ 3 ]

(i)

a value for the Planck constant.

[ 1 ]

(ii)

the work function of the material.

[ 1 ]

(iii)

the threshold wavelength of the material.

[ 1 ]

$\quad \lambda_{0}=\frac{c}{\text { frequency intercept }}$ or $\frac{h c}{\text { max ke intercept }}$ ;

(d)

In an experiment, light at a particular frequency is incident on a surface and electrons are emitted. Explain what happens to the number of electrons emitted per second when the intensity of this light is increased.

[ 2 ]

[Maximum number: 10]

The binding energy of the stable nuclide ${ }_{54}^{131} \mathrm{Xe}$ is 1.105 GeV .

(a)

Outline what is meant by binding energy.

[ 1 ]

(b)

Calculate, in $\mathrm{GeV} \mathrm{c}^{-2}$ , the mass of a nucleus of ${ }_{54}^{131} \mathrm{Xe}$ .

[ 2 ]

Use of Binding energy = mass of separate nucleons - mass of nucleus
$\begin{aligned} \text { « } 1105=54 \times 938+77 \times 940-M c^{2} \text { » } M=122 \text { « } \mathrm{GeVc}^{-2} \text { » } \end{aligned}$

Do not allow BCA.

Award ECF from using binding energy with incorrect sign.

An answer that doesn't include binding energy scores [0].

(c)

$\quad{ }_{54}^{133} \mathrm{Xe}$ and ${ }_{54}^{131} \mathrm{Xe}$ are two isotopes of xenon.

[ 3 ]

(i)

Outline what is meant by isotopes.

[ 2 ]

(ii)

$\quad{ }_{54}^{133} \mathrm{Xe}$ is radioactive. Suggest how the binding energy per nucleon for ${ }_{54}^{131} \mathrm{Xe}$ compares with that of ${ }_{54}^{133} \mathrm{Xe}$ .

[ 1 ]

binding energy per nucleon is a measure of stability so ${ }_{54}^{131} \mathrm{Xe}$ has the greater binding
energy per nucleon

Correct mention of stability required to award the mark.

Marking guidance:

Allow opposite comment for Xe 133.

(d)

The graph shows the variation with time of the activity of a pure sample of ${ }_{54}^{133} \mathrm{Xe}$ .
$A / 10^{7} \mathrm{~Bq}$

[ 4 ]

(i)

Estimate the half-life of ${ }_{54}^{133} \mathrm{Xe}$ .

[ 1 ]

(ii)

Calculate the activity of the sample after 25 days as a fraction of the initial activity.

[ 3 ]

ALTERNATIVE 1

decay constant in range 0.128 to 0.133 «days ${ }^{-1}$ »
$\frac{A}{A_{0}}=e^{-\lambda \times 25}$

answer in range 0.036 to 0.040

ALTERNATIVE 2
$n=\frac{25}{5.3}=\ll 4.7 \text { half lives } \gg$ $\frac{A}{A_{0}}=0.5^{4.7}$

answer in range 0.036 to 0.040

Marking guidance:

Allow ECF from (c)(i)

[Maximum number: 9]

(a)

One possible fission reaction of uranium-235 (U-235) is

{ }_{92}^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \rightarrow{ }_{54}^{140} \mathrm{Xe}+{ }_{38}^{94} \mathrm{Sr}+2{ }_{0}^{1} \mathrm{n}

The following data are available.
Mass of one atom of U-235 =235 u
Binding energy per nucleon for U-235 =7.59 MeV
Binding energy per nucleon for Xe-140 =8.29 MeV
Binding energy per nucleon for Sr-94 =8.59 MeV

[ 2 ]

(i)

State what is meant by binding energy of a nucleus.

[ 1 ]

(ii)

Show that the energy released in the reaction is about 180 MeV .

[ 1 ]

$140 \times 8.29+94 \times 8.59-235 \times 7.59$ OR 184 «MeV»

(b)

A nuclear power station uses U-235 as fuel. Assume that every fission reaction of U-235 gives rise to 180 MeV of energy.

[ 4 ]

(i)

Estimate, in $\mathrm{Jkg}^{-1}$ , the specific energy of U-235.

[ 2 ]

see «energy = » $180 \times 10^{6} \times 1.60 \times 10^{-19}$ AND «mass =» $235 \times 1.66 \times 10^{-27}$ $7.4 \times 10^{13} \ll \mathrm{~J} \mathrm{~kg}^{-1} \gg$

(ii)

The power station has a useful power output of 1.2 GW and an efficiency of 36 %. Determine the mass of U-235 that undergoes fission in one day.

[ 2 ]

$\text { energy produced in one day }=\frac{1.2 \times 10^{9} \times 24 \times 3600}{0.36}=2.9 \times 10^{14} \ll \mathrm{~J}>\text { }$

(c)

A sample of waste produced by the reactor contains 1.0 kg of strontium-94 ( Sr-94 ). Sr-94 is radioactive and undergoes beta-minus ( $\beta^{-}$ ) decay into a daughter nuclide X . The reaction for this decay is

{ }_{38}^{94} \mathrm{Sr} \rightarrow \mathrm{X}+\bar{v}_{e}+e

[ 3 ]

(i)

Write down the proton number of nuclide X .

The graph shows the variation with time of the mass of Sr-94 remaining in the sample.

[ 1 ]

39

Marking guidance:

Do not allow ${ }_{39}^{94} \mathrm{X}$ unless the proton number is indicated.

(ii)

Calculate the mass of Sr -94 remaining in the sample after 10 minutes.

[ 2 ]

ALTERNATIVE 1
$10 \mathrm{~min}=8 t_{1 / 2}$

mass remaining $=1.0 \times\left(\frac{1}{2}\right)^{8}=3.9 \times 10^{-3}<\mathrm{kg} \gg$

ALTERNATIVE 2

decay constant = « $\frac{\ln 2}{75}=$ » $9.24 \times 10^{-3} \ll \mathrm{~s}^{-1}>$ ↓

mass remaining $=1.0 \times e^{-9.24 \times 10^{-3} \times 600}=3.9 \times 10^{-3}$ « kg »

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