Oxford Revise AQA A Level Physics | Chapter P3 answers

P3: Properties of particles

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Answers

Extra information

Marks

AO

Spec. reference

01.1

Energy/momentum was not conserved

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3.2.1.7

01.2

\(\text{n } \to \text{ p}^+ + \text{ e}^- + \bar{\nu}\)

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3.2.1.2

01.3

Lepton

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3.2.1.5

01.4

Charge is conserved in both decays

Baryon number is not conserved in the first decay

Lepton number is not conserved in the second decay

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3.2.1.7

01.5

Proton

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3.2.1.5

01.6

A neutron is a hadron, but the weak interaction is responsible for this reaction/leptons are produced

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3.2.1.5

02.1

\(\text{u } \to \text{ d } + \text{ e}^+ + \nu_{\text{e}}\)

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3.2.1.6

02.2

\(
\begin{aligned}
& \text{Lepton number: }0 = 0 + (-1) + (+1)\\
& \text{Charge: } + \frac{2}{3} = -\frac{1}{3} + (+1) + 0\\
& \text{Baryon number: }\frac{1}{3} = \frac{1}{3} + 0 + 0\\
\end{aligned}
\)

Numbers consistent with equation

All particle numbers must be present in each equation for the mark

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2

3.2.1.7

02.3

Energy/momentum

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3.2.1.7

02.4

\(\mu^+ \to \text{ e}^+ + \nu_{\mathrm{e}} + \bar{\nu}_{\mu}\)

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3.2.1.5

02.5

Two correct points, for example:

Only leptons are produced in antimuon decay

Both decays produce a positron and an electron neutrino

One similarity and one difference

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3.2.1.5

03.1

Correct similarity and difference, for example:

Mesons and baryons are both hadrons/composed of quarks

Mesons contain two quarks/quark-antiquark pairs, but baryons contain three quarks

One similarity and one difference

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1

3.2.1.5

03.2

Protons interact using the strong nuclear force only

Kaons are formed by the strong interaction but decay by the weak interaction

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2

3.2.1.4

03.3

One up and one strange antiquark, or \(\text{u}\bar{\text{s}}\)

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3.2.1.6

03.4

Weak interaction

Strangeness is not conserved/a quark has decayed

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3.2.1.4

03.5

The equipment required is too big and expensive/scientists working in a range of disciplines are needed to produce the reactions

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3.2.1.5

04.1

Proton – two up quarks and one down quark, or uud

Antiproton – two up antiquarks and one down antiquark, or \(\overline{\mathrm{uud}}\)

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3.2.1.6

04.2

The charge on an up quark is \(+ \frac{2}{3}\), on an anti-up quark is \(– \frac{2}{3}\), on a down quark is \(– \frac{1}{3}\), and on an anti-down quark is \(+ \frac{1}{3}\)

\(
\begin{aligned}
& \text{uud – total charge } = \left( +\frac{2}{3} \right) + \left( +\frac{2}{3} \right) + \left( -\frac{1}{3} \right) = +1\\
& \overline{\text{uud}}\ – \text{ total charge } = \left( -\frac{2}{3} \right) + \left( -\frac{2}{3} \right) + \left( +\frac{1}{3} \right) = -1\\
\end{aligned}
\)

Charges on quarks

Two equations showing +1 and –1

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3.2.1.6

04.3

\(
\begin{aligned}
& \text{uud } + \overline{\text{uud}} \to \text{u}\bar{\text{d}} + \text{d}\bar{\text{u}} + \text{u}\bar{\text{u}}\\
& \text{u }\bar{\text{d}} \text{ is a }\pi^+ \text{ meson}\\
& \text{d }\bar{\text{u}} \text{ is a }\pi^- \text{ meson}
\end{aligned}
\)

Evidence of quark structure of products from p and anti-p

\(q\bar{q}\text{ match each particle}\)

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3.2.1.6

04.4

Quarks have a baryon number of \(+ \frac{1}{3}\), antiquarks \(– \frac{1}{3}\)

\(\text{u}\bar{\text{u}}\) annihilate, total baryon number does not change because the total is zero

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3.2.1.6

05.1

The particles are all made up of quarks/there are no leptons produced

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3.2.1.6

05.2

\(\text{p}^+ + \pi^- \to \mathrm{K}^+ + \Sigma^-\)

Negative

Charge before is zero; if kaon is positive, sigma must be negative

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3.2.1.5

05.3

\(
\begin{aligned}
& \text{uud } + \text{d}\bar{\text{u}}\ \to \text{u}\bar{\text{s}} + \text{dds}\\
& \text{LHS:udd}\\
& \text{RHS}: \text{uud } – \text{u}\bar{\text{s}}\ = {\text{dds}}\\
& \text{Charge } = – \frac{1}{3} – \frac{1}{3} – \frac{1}{3} = – 1
\end{aligned}
\)

Evidence of use of quark structure

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3.2.1.6

05.4

The kaon is produced by the strong interaction

But it decays by the weak interaction

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3.2.1.5

06.1

Correct examples, for example:

Hadron: proton, neutron, kaon, pion

Lepton: electron, muon, neutrino

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3.2.1.5

06.2

Hadrons can interact by all four interactions if they are charged (strong, weak, electromagnetic, gravitational)

Leptons do not interact by the strong interaction

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3.2.1.5

06.3

pions are composed of 2 quarks

A pion does not include the proton as a decay product

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3.2.1.5

06.4

\(
\begin{aligned}
& \text{p } + \pi^- \to \pi^0 + \mathrm{\ X}\\
& \text{uud } + \text{d}\bar{\text{u}}\ \to \text{u}\bar{\text{u}} + \text{ udd}\\
& \text{X is a neutron}\\
& \text{Conservation of charge: } + 1 + – 1 \to 0 + 0\\
& \text{Conservation of baryon number: } + 1 + 0 \to 0 + 1
\end{aligned}
\)

Conservation of charge/baryon number/quark number

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3.2.1.7

06.5

Particles have a (de Broglie) wavelength

And are scattered/diffracted if the wavelength is the same order of magnitude as the particles from which it is scattered; here an atom has a diameter ~10–10 m

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3.2.2.4

06.6

\(
\begin{aligned}
\lambda = \frac{h}{p},\ p = \frac{h}{\lambda} & = \frac{6.63 \times 10^{-34}}{10^{-10}}\\
& = 6.63 \times 10^{-24}\ \mathrm{kg\ m\ s}^{-1}
\end{aligned}
\)
\(\text{Use of }p = \frac{h}{\lambda}\)

Answer

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3.2.2.4

07.1

A proton is a charged particle, which experiences a force when it is in an electric field

Neutrons have no charge so cannot be accelerated

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3.7.3.2

07.2

Particle accelerators had not been built/were too expensive to build

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3.2.1.5

07.3

uud

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3.2.1.6

07.4

Both contain quarks/take part in the strong interaction

Mesons contain quark/antiquark pairs, but the proton contains three quarks

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3.2.1.5

07.5

Negative

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3.2.1.7

07.6

X is a baryon

To conserve baryon number: +1 + 0 = 0 + 0 + (+1)

Baryon

Correct equation

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3.2.1.7

07.7

\(
\begin{aligned}
& \text{Strangeness on LHS } = -1 \left( \mathrm{s}\bar{\mathrm{u}} \right)\\
& \text{On right hand side } = +2 \left( \mathrm{d}\bar{\mathrm{s}} + \mathrm{u}\bar{\mathrm{s}} \right)\\
& \text{So X has strangeness of -3}
\end{aligned}
\)

–1 on LHS

Showing –1 = +2 + (–3)

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3.2.1.7

08.1

Mass of nucleus = mass of 26 protons and (56 – 26 =) 30 neutrons

\(
\begin{aligned}
\text{Mass } & = 26 \times 1.673 \times 10^{-27} \ \mathrm{kg} \ + 30 \times 1.675 \times 10^{-27}\ \mathrm{kg}\\
& = 9.38 \times 10^{-26}\ \mathrm{kg}\\
\end{aligned}
\)

Use of A and Z

Answer

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3.2.1.1

08.2

Resolving momentum in the x-direction

\(m_{\text{Fe}}u_{\text{Fe}} = m_{\text{Fe}}v_{\text{Fe,x}} + m_{\text{N}}v_{\text{N,x}}\)

OR

\(m_{\text{Fe}}u_{\text{Fe}} = m_{\text{Fe}}v_{\text{Fe}}\cos 25 + m_{\text{N}}v_{\text{N}}\cos 36\)

OR

\(
\begin{array}{lll}
v_{\text{N}} & = & \frac{m_{\mathrm{Fe}}u_{\mathrm{Fe}} – m_{\mathrm{Fe}}v_{\mathrm{Fe}}\cos 25}{m_{\mathrm{N}}\cos 36}\\
& = & \frac{9.38 \times 10^{-26} \times 7.1 \times 10^7 – 9.38 \times 10^{-26} \times 4.8 \times 10^7 \cos 25}{2.34 \times 10^{-26} \cos 36}\\
& = & 1.38 \times 10^8\ \mathrm{m\ s}^{-1}\\
\end{array}
\)

OR

Resolving momentum in the y-direction

\(m_{\mathrm{Fe}}v_{\mathrm{Fe,y}} = m_{\mathrm{N}}v_{\mathrm{N,y}}\)

OR

\(m_{\mathrm{Fe}}v_{\mathrm{Fe}} \sin 25 = m_{\mathrm{N}}v_{\mathrm{N}}\ \sin \ 36\)

OR

\(
\begin{aligned}
v_{\mathrm{N}} = \frac{m_{\mathrm{Fe}v_{\mathrm{Fe}}\ \sin 25}{m_{\mathrm{N}} \ \sin 36}}\\
\end{aligned}
\)

\(
\begin{aligned}
& = \frac{9.38 \times 10^{-26} \times 4.8 \times 10^7 \ \sin 25}{2.34 \times 10^{-26} \ \sin 36}\\
\end{aligned}
\)

\(= 1.38 \times 10^8\ \mathrm{m\ s}^{-1}\)

Conservation of momentum, explicit or implied

Resolution of momenta in x or y direction

Answer

Answer given to three significant figures

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3.4.1.4

08.3

In an elastic collision, kinetic energy is conserved

Assuming all of the energy of the nuclei was kinetic, if some energy is transferred by gamma rays, there will be less kinetic energy after the collision, so the collision is not elastic

owtte

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3.4.1.6

08.4

\(
\begin{aligned}
\text{Total energy of gamma ray } & = 2 \times 139.567\ \mathrm{MeV}\\
& = 279.134 \times 1.60 \times 10^{-19} \times 10^{6}\\
& = 4.47 \times 10^{-11}\ \mathrm{J}\\
\end{aligned}
\) \(
\begin{aligned}
& E = hf = \frac{hc}{\lambda}\\
& \lambda = \frac{hc}{E}\\
& = 4.45 \times 10^{-15}\ \mathrm{m}
\end{aligned}
\)

Energy calculated

Answer commensurate with energy

ALLOW 4.5 × 10−15

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3.2.1.3

09.1

hf = energy of photon with frequency \(\frac{f}{f}\) is frequency of photon/light and h is the Planks constant

ϕ = work function / energy required to remove an electron from the surface of a metal

Ek= kinetic energy of (ejected) electron

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3.2.2.1

09.2

\(\text{Work function } = 5.01 \times 1.6 \times 10^{-19}\ \mathrm{J} = 8.016 \times 10^{-19} \mathrm{J}\) \(E = hf,\ f = \frac{E}{h} = \frac{8.016 \times 10^{-19}}{6.63 \times 10^{-34}} = 1.21 \times 10^{15}\ \mathrm{Hz}\)

Energy in joules

Substitution

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3.2.2.1

09.3

\(
\begin{aligned}
\text{Difference in energy } = hf – \phi & = 4.2 \times 10^{15} \times 6.63 \times 10^{-34} – 8.016 \times 10^{-19}\ \mathrm{J}\\
& = 1.98 \times 10^{-18}\ \mathrm{J}\\
& = \frac{1}{2}\ mv^2\\
\end{aligned}
\) \(
\begin{aligned}
v & = \sqrt{\frac{2 \times 1.98 \times 10^{-19}}{9.11 \times 10^{-31}}}\\
& = 2.09 \times 10^6\ \mathrm{m\ s}^{-1}
\end{aligned}
\)

Energy

Substitution

Answer

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3.2.2.1

09.4

\(
\begin{aligned}
\lambda = \frac{h}{mv} & = \frac{6.63 \times 10^{-34}}{9.11 \times 10^{-31} \times 2.09 \times 10^6}\\
& = 3.48 \times 10^{-10}\ \mathrm{m}
\end{aligned}
\)

Substitution

Answer

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3.2.2.4

09.5

Yes, the wavelength is the same as the order of magnitude as the spacing of atoms

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3.2.2.4

09.6

Decreasing the p.d. decreases the momentum of the electrons

Increases the de Broglie wavelength of the electrons

Increases the diameter of the rings

Decreases the brightness of the rings

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3.2.2.4

10.1

\(hf = \phi + {E_{\mathrm{k}\left( \mathrm{max} \right)}}\)

\(E_{\mathrm{k}\left( \mathrm{max} \right)} = e{V_s}\)

\(
\begin{aligned}
\phi & = h\,f – e\,V_s\\
\phi & = \frac{h\,c}{\lambda} – e\,{V_s}\\
& = \frac{6.63 \times 10^{-34} \mathrm{\ J\ s\ } \times 3.0 \times 10^8 \mathrm{\ m\ s}^{-1}}{188 \times 10^{-9} \mathrm{\ m}} – 4.5 \times 1.6 \times 10^{-19}\\
& = 3.38 \times 10^{-19}\ \mathrm{J}\\
& = 2.11\ \mathrm{eV}\\
\end{aligned}
\)

The cathode is made of caesium

Combining equations

Work function in J/eV

Metal consistent with answer

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3.2.2.1

3.2.2.2

10.2

If a metal with a higher work function is used the energy required to remove the electron will be greater

The kinetic energy of the photoelectron will be smaller, so the stopping potential will be smaller

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3.2.2.1

10.3

The radiation promotes an electron to a higher energy level

When the electron moves back to its original state it emits a photon of visible light

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3.2.2.3

10.4

In the scintillator particles collide with the chemical on the screen

In the fluorescent tube UV light emitted by mercury atoms is incident on the coating on the inside of the tube

In both cases electrons in the atoms of screen / coating are promoted to a higher energy level

And emit visible light when they move back down

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3.2.2.2

3.2.2.3

10.5

An electron moves from level C to level B when the atom absorbs a photon

When an electron moves from level B to level C a photon is emitted

The energy / frequency of the photon is the same in each case

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3.2.2.3

10.6

Energy of photon = 1.8 × 1.6 × 10−19 J

= 2.89 × 10−19 J

ΔE= (−8.86 × 10−19 J – E) = 2.89 × 10−19 J

E = −5.97 × 10−19 J

Energy of photon

Substitution

Correct value of energy

Negative value

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3.2.2.3

10.7

\(
\begin{aligned}
\lambda = \frac{h\,c}{E} & = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{2.89 \times 10^{-19}}\\
& = 6.87 \times 10^{-7}\ \mathrm{m}\ /\ 687\ \mathrm{nm}\\
\end{aligned}
\)

OR

\(f = \frac{E}{h} = \frac{2.89 \times 10^{-19}}{6.63 \times 10^{-34}} = 4.36 \times 10^{14}\mathrm{\ Hz}\)

No, this is in the visible region / red of the electromagnetic spectrum

Substitution

Answer

Conclusion

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3.2.2.4

Skills box answers

Question

Answer

1

\(
\begin{aligned}
& hf = E_{\mathrm{k}} + \phi,\text{ therefore } f = \frac{\left( E_{\mathrm{k}} + \phi \right)}{h}\\
& E_k = 1.1\ \mathrm{eV} \times 1.6 \times 10^{-19}\ \mathrm{J}\ \mathrm{eV}^{-1} = 1.76 \times 10^{-19}\ \mathrm{J}\\
& \phi = 4.08\ \mathrm{eV} \times 1.6 \times 10^{-19}\ \mathrm{J}\ \mathrm{eV}^{-1} = 6.528 \times 10^{-19}\ \mathrm{J}\\
& f = \frac{\left( 1.76 + 6.528 \right) \times 10^{-19}\ \mathrm{J}}{6.63 \times 10^{-34}\ \mathrm{J}\ \mathrm{s}} = 1.3 \times 10^{15}\ \mathrm{Hz}
\end{aligned}
\)

2

hf = Ek (max) + ϕ

but in this case, Ek (max) = 0

ϕ = 2.87 × 1.6 × 10–19 J = 4.59 × 10–19 J

therefore threshold frequency = \(\frac{4.59 \times 10^{-19}\ \mathrm{J}} {6.63 \times 10^{-34}\ \mathrm{J\ s}} = 6.9 \times 10^{14}\ \mathrm{Hz}\)

3

Rearrange the equation to make ϕ the subject: ϕ = hfEk (max)

Ek (max) = 1.3 × 1.6 × 10–19 J = 2.08 × 10–19 J.

Photon energy hf = 3.2 × 10–19 J

ϕ = 3.2 × 10–19 J – 2.08 × 10–19 J = 1.12 × 10–19 J

To convert to eV, divide by 1.6 × 10–19 J eV–1:

ϕ = 0.7 eV

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