Oxford Revise AQA GCSE Physics | Chapter P7 answers

P7: Atoms

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Answers

Extra information

Mark

AO / Specification reference

01.1

10

1

AO2

4.4.1.2

01.2

the number of electrons is the same as the number of protons

electrons and protons have opposite charges/charge cancels/atoms are neutral so have equal positive and negative charges

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1

AO1

AO2

4.4.1.1

01.3

A and B

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AO1

4.4.1.1

01.4

isotopes have the same number of protons

but different numbers of neutrons

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1

AO1

4.4.1.2

02.1

particle B

there are the same number of those particles in each nucleus

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1

AO1

4.4.1.2

02.2

particle A is a neutron

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AO1

4.4.1.2

02.3

the element is lithium

it has an atomic number of three

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1

AO1

4.4.1.2

02.4

\(\textrm{isotope one: }{}_{3}^{6}{\rm{Li}}\)

\(\textrm{isotope two: }{}_{3}^{7}{\rm{Li}}\)

one mark for correct top number/atomic mass in each case

one mark for correct bottom number/atomic number in each case

one mark for correct symbol (credit once only)

5

AO1

4.4.1.2

02.5

the same

the charge comes from/depends on the number of protons, which is the same

1

1

AO1

4.4.1.2

03.1

10–10 m

1

AO1

4.4.1.1

03.2

an orbit/distance from the nucleus where there are atoms

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AO1

4.4.1.1

03.3

when hydrogen absorbs electromagnetic radiation, the electron will move up a level

e.g., from level one to level two or three

when the electron moves back down it emits electromagnetic radiation

e.g., when it moves from three to two, or from two to one

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1

1

1

AO1

AO1

4.4.1.1

04.1

looking at the atomic/bottom number/seven

subtracting the atomic/bottom number from the mass/top number

looking at the atomic/bottom number/seven/it is the same as the number of protons

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1

1

AO1

4.4.1.2

04.2

no, there are atoms where the mass/top number/atomic mass is not double the atomic/bottom number/atomic number

suitable example e.g., Be, Na

1

1

AO1

AO1

4.4.1.2

05

Level 3: Description of the how the nuclear model came to be proposed, and the limitations of it. Comment about the refinement of the model when the proton and neutron were discovered.

5-6

AO1

AO1

4.4.1.3

Level 2: Description of the nuclear model and some discussion about how the structure of the nucleus wasn’t worked out until later.

3-4

Level 1: Some comment about the development of the model in terms of the alpha scattering experiment and/or discovery of particles in the nucleus.

1-2

No relevant comment.

0

Indicative content:

  • the discovery of the electron was before the nuclear model was proposed
  • Rutherford knew that electrons were negatively charged and that atoms were neutral
  • so he knew that there had to be positive charge in the nucleus
  • the results of the alpha particle experiment showed that there was a massive, small positively charged object at the centre of the atom
  • and that most of the atom was empty space
  • his model included a positive massive nucleus but did not include the names or number of particles in it

the model was refined when the proton and neutron were discovered and named

06.1

some energy from the gravitational potential energy store at the start is transferred to the thermal energy store of the surroundings when the ball bounces

so there is less energy in the gravitational potential energy store at the end of the bounce

1

AO1

AO2

4.1.1.1

06.2

points should be plotted at:

(20, 16), (40, 30), (60, 49), (80, 55), (100, 70)

one mark for correct variables on the x and y-axes

one mark for appropriate scales on the x and y-axes with labels and units

one mark for plotting three or four points of data correctly (within ± 0.5 squares)

two marks for plotting all points of data correctly

one mark for drawing an acceptable line of best fit

5

AO2

AO3

06.3

the anomalous results is for the 60 cm drop/49 cm bounce

1

AO3

06.4

yes/bounce height is proportional to the drop height

the line of best fit is a straight line through (0, 0)/origin

or as drop height doubles, the bounce height doubles

1

1

AO3

07.1

\(
\begin{array}{lr}
\textrm{percentage} = \frac{33 + 43 + 52 + 70}{142\;141} \times 100\ \%\\
= \frac{198}{142\;141} \times 100\ \%\\
= 0.14\ \%
\end{array}
\)

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AO2

4.4.1.3

07.2

percentage going through = 100 – 0.14 = 99.86 %

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AO2

4.4.1.3

07.3

the majority of particles went through the foil

so the part of the atom/nucleus deflecting alpha particles was very small.

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1

AO3

4.4.1.3

07.4

positive (or 2+)

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AO1

4.4.1.3

07.5

If the nucleus is negative the alpha particles would be attracted to the nucleus and there would be no particles repelled and no deflections greater than 90°

If it is positive, they would be repelled

and be deflected by small or larger angles

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1

1

1

AO1

AO3

4.4.1.3

08.1

A, D, E, F

two marks for three or four correct

one mark for one or two correct

2

AO1

4.4.1.1

08.2

B

C

1

1

AO1

4.3.3.2

08.3

E

F

1

1

AO1

4.4.1.1

08.4

positive (2+)

1

AO3

4.4.1.2

08.5

the nucleus is smaller by a factor of

\(
\begin{array}{lr}
\textrm{so the nucleus should be}\ \frac{0.2}{10\;000}\\
= 2 \times 10^{-6}\ \textrm{m}\\
\textrm{this is}\ \frac{1}{1000}\ \text{of a millimetre/too small to draw}
\end{array}
\)

1

1

1

1

AO1

4.4.1.1

09.1

the plum pudding model is a model with a positively charged mass

with negatively charged electrons

embedded in it

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1

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AO1

4.4.1.3

09.2

the plum pudding model replaced the Dalton model, which was that atoms were tiny spheres that could not be divided

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AO1

4.4.1.3

09.3

most alpha particles went straight through a gold foil

but some came back/were deflected through more than 90°

1

1

AO1

4.4.1.3

10.1

if the current in the circuit becomes too large the wire inside a fuse melts

which breaks the circuit (and helps to prevent injury)

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1

AO1

4.2.3.2

10.2

it takes a certain amount of energy to raise the temperature of the metal sufficiently to melt the fuse wire

the energy required depends on the mass of the wire/thinner wires can only carry smaller currents

if the wire needs to melt at a higher current the mass needs to be bigger, so the wire needs to be thicker

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1

1

AO1

AO2

4.3.2.2

10.3

mass = density × volume

1

AO1

10.4

0.5 cm = 5 × 10–3 m

volume = length × area

= 5 × 10–3 × 1 × 10–6

= 5 × 10–9

mass = 7000 × 5 × 10–9

= 3.5 × 10–5 kg

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1

1

1

1

AO1

AO2

10.5

energy = mass × specific heat capacity × change in temperature

= 3.5 × 10–5 × 230 × (687 – 20)

= 5.37 J (5.4 J)

energy = mass × specific latent heat

= 3.5 × 10–5 × 300 000

= 10.5 J

Total energy = 5.37 + 10.5 = 15.87 = 15.9 J

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1

1

1

AO2

4.3.1.1

4.3.2.2

4.3.2.3

10.6

energy transferred = power × time

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AO1

10.7

power = current2 × resistance

1

AO1

10.8

15.87 = current2 × 1.8 × 0.5

\(
\begin{array}{lr}
\textrm{current = }\sqrt {\frac{15.87}{1.8 \times 0.5}} \\
= 4.20\ \rm{A}
\end{array}
\)

or

\(
\begin{array}{lr}
\textrm{P} = \frac{\rm{E}}{\rm{t}} = \frac{15.87}{0.5}
= 31.74\ \rm{W}\\
\rm{I}^2 = \frac{\rm{P}}{\rm{R}} = \frac{31.74}{1.8}= 17.63\\
\rm{I} = \sqrt{17.63} = 4.2\ \rm{A}
\end{array}
\)

allow E = I2Rt

accept 4.2 with no working shown for two marks

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1

1

or

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1

1

AO2

4.2.4.2

11.1

one mark for battery, ammeter and component in series

one mark for voltmeter in parallel with the component

one mark for battery, ammeter and voltmeter correctly labelled

3

AO2

4.2.1.4

11.2

(connect up the circuit, and) measure the potential difference and current record the measurements

change the potential difference of the battery/add a variable resistor, and take another set of measurements

repeat for different values of potential difference

calculate the ratio of potential difference to current/plot a graph of potential difference against current.

if the ratio/gradient is constant and straight line goes through origin is constant, the component is ohmic

1

1

1

1

AO1

AO1

4.2.1.4

11.3

\(
\begin{array}{lr}
\textrm{gradient of this graph}= \frac{1}{\rm{V}}\ \textrm{which equals}\ \frac{1}{\rm{R}}\\
\textbf{or}\ \rm{R} = \frac{1}{\rm{gradient}}
\end{array}
\)

1

AO1

AO1

4.2.1.4

11.4

one mark for drawing the graph in third quadrant

one mark for curved lines

one mark for correct curvature

3

AO3

4.2.1.4

11.5

potential difference = current × resistance

1

AO1

4.2.1.3

11.6

\(
\begin{array}{lr}
\rm{resistance} = \frac{\rm{V}}{\rm{I}}\\
= \frac{6}{0.6}\\
= 10\ \Omega
\end{array}
\)

1

1

AO1

4.2.1.3

11.7

the resistance changes by a factor of two (is not constant)/not a straight line through the origin

so the component is non-ohmic

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1

AO3

4.2.1.4

12.1

greater than

greater than

same as

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1

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AO1

AO1

4.3.2.1

12.2

greater than

1

4.3.2.1

12.3

the specific heat capacity for hot water is the same as for cold water

1

4.3.2.2

12.4

the average speed and specific heat capacity will stay the same

because these quantities do not depend on the number of particles in the beaker.

the kinetic energy, potential energy and internal energy will be halved

because the number of particles in the beaker has halved.

1

1

1

1

AO1

AO1

AO2

4.3.2.1

13

Level 3: Clear and coherent description and comparison of fission and fusion.

5-6

AO1

AO1

4.4.4.1

4.4.4.2

Level 2: Processes of fission and fusion described but not compared.

3-4

Level 1: A relevant part or description of fission or fusion with no comparison.

1-2

No relevant comment.

0

Indicative content:

  • fission is the splitting of large unstable nuclei
  • into smaller nuclei
  • fusion is the joining together of smaller nuclei
  • to make larger nuclei
  • fission processes can be started and maintained easily
  • fission occurs at lower temperatures / fusion occurs at higher temperatures
  • fission is a process that happens in nuclear power stations but there are no nuclear fusion power stations
  • fission causes a chain reaction/fusion does not
  • both release energy (more energy released in fusion)

14.1

a positively charged mass

with negatively charged electrons embedded in it

1

1

AO1

4.4.1.3

14.2

two correct points e.g.,

the electron was smaller/less massive than the atom

it showed that there were particles that were smaller than the atom

it was negatively charged showing that there was positive and negative charge in the atom

one mark for each correct point up to a maximum of two marks

2

AO1

AO2

4.4.1.3

14.3

all of the alpha particles would go through or all the alpha particles would come back

1

AO1

4.4.1.3

14.4

some of the alpha particle came back, but most went through

which could not be explained by the current model, so the model had to change

1

AO2

4.4.1.3

14.5

correct suggestion e.g.,

the scientists were using the results to develop an alternative model

the results were being checked by other scientists

1

AO2

4.4.1.3

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