AS Physics unit 1 AQA June 13 unofficial markscheme

Can anyone remember the marks for 3bi? Thank you for posting these!

Number 1
a) 20 protons , 28 neutrons and 18 electrons ( 3 marks)
bi) charge = 2+ = 3.2*10^-19 (1 mark)
ii) specific charge = part bi divided by mass of ion = 4.0*10^6Ckg^-1 (2 marks)

Number 2
a) A = down quark (d) , B = W+ meson , C = positron (e+) and D = neutrino (4 marks)
bi) weak interaction (1 mark)
ii)exchange particle = B (1 mark)
iii) difference between B and photon - photon has no charge or mass and particle B does (1 mark)
c) 2 things conserved in this process - any 2:
charge - charge before = charge after = +2/3
lepton number - before = 0 = after the decay
strangeness - before = 0 = to after the decay
baryon number - before = 1/3 after = 1/3
(4 marks)

Number 3
a) hadrons interat via stong and weak, lepton only via weak
hadrons split into 2 groups - baryons ( 3 quark or 3 anti quark structure ) and mesons ( quark anti quark structure)
example of lepton = electrons etc and hadrons = proton ( baryon aswell) and pion ( meson aswell)
similarity is that both groups of particles interact via weak force and both groups include anti particles
extra points - hadrons not fundemental , leptons are
(6 marks )
bi) any particle anti particle pair (1 or 2 marks not sure ? )
ii) these particle anti particles differ - OPPOSITE charges ( 1 mark)

Number 4
a) Energy of photon = hf , 1 to 1 interaction with electrons , kinetic energies vary up to a certain maximum as electrons position vary so the energy required to remove the electron varies . Those electrons at the surface are emitted with max Ek as the energy required to remove them = work functiom. Mention of formula aswell maybe. Range of ke's because more energy needed to remove deeper electrons etc.(3 marks)
bi) 2.89*10^-19J = 1.8eV = work function (3 marks )
ii) 2.89*10^-19 / h constant = 4.4 * 10^14 Hz (any 2 sf answer gives 1 mark ) (3 marks)
ci) frequency decreases , energy of photon decreases as E = hf and is dependent upon f , max Ek decreases of emitted electrons (2 marks )
ii) Intensity doubled , double the photons incident on the metal per second , double the electrons emitted per second (2 marks)

Number 5
ai) peak to peak = 128 V (1 mark)
ii) peak = 128/2 = 64V (1 mark)
iii)rms = 64/root2 = 45V (2 marks)
bi) 1 cycle = 10 ms , T = 10ms = 10*10^-3 s , f=1/T = 100 Hz ( 3 marks , 1 mark for Hz)
ii) a horizontal line at 45 V on figure 2 ( + or - 45 V i think) (2 marks)
c) Initially have time base off, u will get a vertical line = peak to peak voltage = 128V. 8 boxes vertically so set the y gain to around 128/8 = 16V div ^-1. We need 2 cycles, 1 cycle = 10ms 2 = 20 ms and horizontally we have 10 boxes so 20/10 = 2ms div^-1 = 2*10^-3s div^-1 ( 3 marks )

Number 6
ai)4.2*1.5 = 6.3 V (1 mark)
ii)12-6.3 = 5.7 V ( 1 mark)
iii) current through 2 ohm = 5.7/2 =2.85 A (1 mark)
iv) current through R = 4.2-2.85 = 1.35A (1 mark)
v) resistance of R = 5.7/1.35 = 4.22 ohms (1 mark)
vi)overall resistance of circuit = 2.9 ohms (2 marks)
b)26.46 W
16.245 W
7.695 W (3 marks)
c) energy is conserved as power is conserved. Power in batter = 12*4.2 = 50.4W and the total power of part b = 50.4 W (2 marks)

Number 7
a) current = V/R = 6/90,000 = 6.67*10^-5A (1mark)
ii) voltmeter reads 5000 * 6.67*10^-5 = 0.33 V (1 mark)
b) As light on LDR increases , resistance decreases , overall resistance decreases , current increases , V=IR so voltmeter reading increases (2 marks)
c) across 5000 ohm resistor voltage = 0.75, current = 0.75/5000 = 0.00015A total circuit current. Total voltage / total resistance = total circuit current , 6/total resistance = 0.00015, total resistance = 6/0.00015 = 40,000 ohms and 2 resistors each = 5000 ohms so 40,000 - 10,000 = 30, 000 ohms ( 3marks)

Number 1
a) 20 protons , 28 neutrons and 18 electrons ( 3 marks)
bi) charge = 2+ = 3.2*10^-19 (1 mark)
ii) specific charge = part bi divided by mass of ion = 4.0*10^6Ckg^-1 (2 marks)

Number 2
a) A = down quark (d) , B = W+ meson , C = positron (e+) and D = neutrino (4 marks)
bi) weak interaction (1 mark)
ii)exchange particle = B (1 mark)
iii) difference between B and photon - photon has no charge or mass and particle B does (1 mark)
c) 2 things conserved in this process - any 2:
charge - charge before = charge after = +2/3
lepton number - before = 0 = after the decay
strangeness - before = 0 = to after the decay
baryon number - before = 1/3 after = 1/3
(4 marks)

Number 3
a) hadrons interat via stong and weak, lepton only via weak
hadrons split into 2 groups - baryons ( 3 quark or 3 anti quark structure ) and mesons ( quark anti quark structure)
example of lepton = electrons etc and hadrons = proton ( baryon aswell) and pion ( meson aswell)
similarity is that both groups of particles interact via weak force and both groups include anti particles
extra points - hadrons not fundemental , leptons are
(6 marks )
bi) any particle anti particle pair (1 or 2 marks not sure ? )
ii) these particle anti particles differ - OPPOSITE charges ( 1 mark)

Number 4
a) Energy of photon = hf , 1 to 1 interaction with electrons , kinetic energies vary up to a certain maximum as electrons position vary so the energy required to remove the electron varies . Those electrons at the surface are emitted with max Ek as the energy required to remove them = work functiom. Mention of formula aswell maybe. Range of ke's because more energy needed to remove deeper electrons etc.(3 marks)
bi) 2.89*10^-19J = 1.8eV = work function (3 marks )
ii) 2.89*10^-19 / h constant = 4.4 * 10^14 Hz (any 2 sf answer gives 1 mark ) (3 marks)
ci) frequency decreases , energy of photon decreases as E = hf and is dependent upon f , max Ek decreases of emitted electrons (2 marks )
ii) Intensity doubled , double the photons incident on the metal per second , double the electrons emitted per second (2 marks)

Number 5
ai) peak to peak = 128 V (1 mark)
ii) peak = 128/2 = 64V (1 mark)
iii)rms = 64/root2 = 45V (2 marks)
bi) 1 cycle = 10 ms , T = 10ms = 10*10^-3 s , f=1/T = 100 Hz ( 3 marks , 1 mark for Hz)
ii) a horizontal line at 45 V on figure 2 ( + or - 45 V i think) (2 marks)
c) Initially have time base off, u will get a vertical line = peak to peak voltage = 128V. 8 boxes vertically so set the y gain to around 128/8 = 16V div ^-1. We need 2 cycles, 1 cycle = 10ms 2 = 20 ms and horizontally we have 10 boxes so 20/10 = 2ms div^-1 = 2*10^-3s div^-1 ( 3 marks )

Number 6
ai)4.2*1.5 = 6.3 V (1 mark)
ii)12-6.3 = 5.7 V ( 1 mark)
iii) current through 2 ohm = 5.7/2 =2.85 A (1 mark)
iv) current through R = 4.2-2.85 = 1.35A (1 mark)
v) resistance of R = 5.7/1.35 = 4.22 ohms (1 mark)
vi)overall resistance of circuit = 2.9 ohms (2 marks)
b)26.46 W
16.245 W
7.695 W (3 marks)
c) energy is conserved as power is conserved. Power in batter = 12*4.2 = 50.4W and the total power of part b = 50.4 W (2 marks)

Number 7
a) current = V/R = 6/90,000 = 6.67*10^-5A (1mark)
ii) voltmeter reads 5000 * 6.67*10^-5 = 0.33 V (1 mark)
b) As light on LDR increases , resistance decreases , overall resistance decreases , current increases , V=IR so voltmeter reading increases (2 marks)
c) across 5000 ohm resistor voltage = 0.75, current = 0.75/5000 = 0.00015A total circuit current. Total voltage / total resistance = total circuit current , 6/total resistance = 0.00015, total resistance = 6/0.00015 = 40,000 ohms and 2 resistors each = 5000 ohms so 40,000 - 10,000 = 30, 000 ohms ( 3marks)

The question was, name a particle and its antiparticle. (1 mark) eg electron positron
The second part was, state a difference between a particle and antiparticle. (1 mark) eg opposite charge

For 2 part C do you have to show that how they are conserved or just state it

is neutron okay instead of d quark for Q2, as it did say which particle do these letters represent

I carried through the rounded values and used P=v²/r and used my rounded value of 2.9 for the total resistance. I ended up getting concordant results but of 49.7 rather than 50.4. (7.7+16+26 = 49.7 144/2.9= 49.7) I think it should only be 1 out of 2 hopefully.

i put baryon number is conserved as it is +1 before and +1 after rather than the 1/3 is this right? and also i just put d. No quark or anything after, is this ok?

Just looking to clarify a few things:

2b) I wrote that the photon is virtual and cannot be detected whereas the weak boson can.

2c) I wrote about the particle as a whole, saying that the charge from a proton (+1) is preserved with the charge of the positron, would that be okay?

4cii) I wrote that the kinetic energy of the electrons would decrease until it was below the threshold frequency where no electrons would then be emitted?

6b) For the values in the table is it okay that I rounded to 1 dp?

7b) If i accidentally read the question as the light intensity decreasing but explained it correctly, would what happen then?

Thanks

For 6 c I got 50.3then some numbers, however I put them to 2 sig figs so both equaled 50


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Did 6b) say not to round the answers?

also i assume 20v per division would be okay for 5c) ?

is neutron okay instead of d quark for Q2, as it did say which particle do these letters represent

Just looking to clarify a few things:

2b) I wrote that the photon is virtual and cannot be detected whereas the weak boson can.

2c) I wrote about the particle as a whole, saying that the charge from a proton (+1) is preserved with the charge of the positron, would that be okay?

4cii) I wrote that the kinetic energy of the electrons would decrease until it was below the threshold frequency where no electrons would then be emitted?

6b) For the values in the table is it okay that I rounded to 1 dp?

7b) If i accidentally read the question as the light intensity decreasing but explained it correctly, would what happen then?

Thanks