Oxford Revise AQA A Level Physics | Chapter P2 answers

Question

Answers

Extra information

Marks

AO

Spec reference

01.1

55 protons

55 electrons

82 neutrons

1

2

3.2.1.1

01.2

Antineutrino

56

Correct β⁻ symbol
\(
\left( {{}_{-1}^0{\rm{e}}} \right)
\)

1

2

3.2.1.2

01.3

Weak interaction

1

1

3.2.1.3

01.4

The stable nucleus contains four more neutrons

AND The same number of protons

owtte

1

3

3.2.1.1

01.5

133

The mass does not change/the electron has no mass

1

2

3.2.1.2

01.6

In β⁻ decay, a neutron changes to a proton

In electron capture, a proton changes to a neutron

1

1

3.2.1.2

02.1

The carbon atoms have different numbers of neutrons AND same number of protons

1

1

3.2.1.2

02.2

\(
\begin{array}{lll}
\rm{Charge} = +6\rm{e} & = & 6 \times 1.60 \times 10^{-19}\ \rm{C}\\
& = & 9.60 \times 10^{-19}\ \rm{C}\\
\frac{\rm{charge}}{\rm{mass}}\ & = & \frac{{9.60 \times {{10}^{-19}}\,{\rm{C}}}}{{2.009 \times {{10}^{-27}}\,{\rm{kg}}}}\\
& = & 4.779 \times 10^8\ \textrm{C kg}^{-1}
\end{array}
\)

1

2

3.2.1.1

02.3

Scale on x-axis showing 1 fm per square

Graph showing repulsive force below about 0.5 fm

Attractive force (below axis) peaking at about 1 fm

Tending to zero beyond 3 fm

As nucleons are pushed together, the strong nuclear force maintains their distance.

1

1

3.2.1.2

02.4

6 positrons AND 6 antiprotons AND 6 antineutrons

1

2

3.2.1.3

02.5

No difference

The difference between matter and antimatter is charge

The strong nuclear force is charge independent

1

3

3.2.1.3

03.1

Similarity: the atomic number changes

Difference: the mass number only changes in alpha decay

1

1

3.2.1.2

03.2

Atomic number is reduced by 6

Each alpha particle reduces the atomic number by 2, so five reduces it by 10

Difference is −4, so 4 β⁻ particles are emitted

1

2

3.2.1.2

03.3

The energy of alpha particles emitted from a radioactive element was approximately the same

The energy of beta particles could vary from very small up to a maximum

The difference in energy between the maximum and that of the beta particle was hypothesised to be carried by a particle called a neutrino

1

1

3.2.1.2

03.4

Neutrinos have no charge

And a very small/negligible mass

1

1

3.2.1.2

04.1

They have different charges

But the same mass

1

1

3.2.1.3

04.2

Total energy = 0.51 × 2 × 1.6 × 10⁻¹³ J

= 1.64 × 10^–13 J

Energy of one photon = 0.82 × 10⁻¹³ J

Use of twice rest mass of electron

1

2

3.2.1.3

04.3

The frequency will be higher

The rest masses of the proton and antiproton are higher, so the energy is higher

1

2

3.2.1.3

04.4

Total energy = 2 × 938.257 MeV × 1.6 × 10⁻¹³ J MeV⁻¹

= 3.002 × 10⁻⁹ J

2

3.2.1.3

05.1

The electromagnetic interaction is a result of the exchange of virtual photons

The ball represents the photon exchanged by two charged particles that are repelling each other

The model does not work so well for particles that attract each other

1

3

3.2.1.4

05.2

Strong (nuclear) interaction

Weak (nuclear) interaction

1

1

3.2.1.4

05.3

W⁻

1

1

3.2.1.4

05.4

p + e → n + ν

The W boson is a virtual particle that mediates the force/decay

1

2

3.2.1.4

05.5

The range of the W boson is very short

The range of the photon is infinite

1

2

3.2.1.4

06.1

24 – 11 = 13 neutrons

1

2

3.2.1.1

06.2

Mass of nucleus = 11 × 1.673 × 10⁻²⁷ kg + 13 × 1.675 × 10^–27 kg

Calculation of mass

Calculation of charge

Specific charge

1

2

3.2.1.1

06.3

An alpha particle contains 2 protons and 2 neutrons,

so must contain 11 + 2 = 13 protons, and 13 + 2 = 15 neutrons

Both n and p numbers correct

Both A and Z correct

1

3

3.2.1.2

06.4

The electrostatic force is repulsive between two protons, but does not affect neutrons

The strong nuclear force between two protons and between two neutrons is the same

The electrostatic force decreases with distance, but is infinite in range

The strong nuclear force is constant, but has a very short (fm) range

1

3

3.2.1.4

06.5

Correct A and Z

antineutrino

1

2

3.2.1.2

06.6

1

2

3.2.1.3

06.7

Both emissions are part of the electromagnetic spectrum

Both emissions are a result of the atom returning from an excited state to lower energy/ground state

The energy of the excited state of the nucleus is much higher, so the frequency of the radiation emitted is in the gamma ray region of the electromagnetic spectrum

Whereas emission of visible light is due to the transition of electrons from high to low energy levels

1

3

07.1

88 protons

138 neutrons

1

2

3.2.1.1

07.2

Symbol for alpha / Allow He

A and Z for Rn

1

2

3.2.1.2

07.3

1

2

3.2.1.3

07.4

Suggested mechanism, such as:

The alpha particle collides with an atom in the paint

An electron is excited to a higher energy level, and emits a photon when it returns to its ground state

Collision producing excitation

Emission of photon

3

07.5

The mechanism involves energy levels, for which the evidence is line spectra/the line spectra of hydrogen

1

1

3.2.2.3

08.1

Mass of electron/positron = 0.51 MeV

Total energy of annihilation = 2 × 0.51 MeV = 1.02 MeV = 1.632 × 10⁻¹³ J

Use of energy of one particle

Frequency

Wavelength

2

3.2.1.3

08.2

Conservation of momentum

Because no other particles are produced in the annihilation

1

2

3.2.1.7

08.3

Leptons

1

1

3.2.1.5

08.4

Leptons are fundamental particles

Mesons are made of quarks, which are fundamental particles

1

1

3.2.1.5

08.5

Evidence of use of quark structure

1

2

3.2.1.6

08.6

Evidence of quark structure equation

equation

1

2

3.2.1.6

Question

Answer

1(a)

2 s.f.

1(b)

3 s.f.

1(c)

3 s.f.

2(a)

2(b)

2(c)

Using E = h f (h is given in data table = 6.63 × 10^–34 J s): E = 6.63 × 10^–34 J s × 750 × 10¹² Hz = 4.97 × 10^–19 J (3 s.f.)