# Physics A G482 EWP Unofficial mark Scheme

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OCR Physics A G482 EWP 9th June 2014

Usual disclaimers.

These are just my answers and are in no sense 'official'.

They may contain errors / typos.

I think this paper was one the hardest I've seen for a long time. Lots of tricky bits.

Expecting low grade boundaries.

Q1 a) i) Not a straight line through origin

so I is not prop to V

so doesn't obey Ohm's law [1]

ii) Line is approx straight so R is approx constant

R = V / I = 0.5v / 50mA = 10ohms [2]

iii) At 6v filament gets hot ( more power dissipated)

Atoms oscillate with bigger amplitude so more collisions with electrons so more energy converted to heat

so resistance increases [2]

b) In the daytime Rldr = 1ohm

so Rcombination is approx 1 ohm

so Pd divides 25:1 with resistor getting 25/26 of 12v and LDR getting 1/26 of 12v

PD across light bulb = PD across LDR is small and doesnt light

At night Rldr = 1000ohm

so R combination = R light bulb approx = 25ohm

so light bulb gets 1/2 of 12v = 6v

so lights up. [5]

Tricky potential divider with that parallel combination.

TOTAL 10

Q2 a) Emf = J/C; resistance = V/A; energy = VC; charge = As [2]

b) i) PD is the energy converted per unit charge from electrical to other forms of energy in a load. [2]

ii) Internal resistance = resistance between the terminals of the cell. [1]

iii) Same PD (in parallel) I = V/R so if half the resistance in branch (6ohm and 12 ohm)) then it

must have twice the current [2]

iv) PD across top branch = I x total R = 0.16x6 = 0.96v

so 0.96v across bottom branch

Potential divider so PD across 6ohm = 6/12 x 0.96 = 0.48v

[A lot of work for 2 marks]

v) PD splits 3:3 in top branch and 6:6 in bottom branch

same potential at each point so no difference [2]

vi) Current in bottom branch = 0.08A (half the other one)

so current through cell = 0.16 + 0.08 = 0.24A

V = E - Ir so 0.96 = 1.2 - 0.24 r

so r= 1.0ohm (possible -1 for sf of 1ohm answer) [4]

A tough question after the easy definitions

TOTAL 15

Q3 a) resistance = PD / Current [1]

b) i) strange setup which will throw many

R = rho x L / A = 1.7E-8 x 20 x 3.8E-10 / (4 x 3.8E-10) ^2

= 224 ohm [3]

ii) no density of electrons = no of atoms per unit vol

= 4 x 20 x 1 / ( 4d x d x 20 d) = 1/ 3.8E-10^3 = 1.82E28 /m3 [1]

Hard work for 1 mark

iii) I = nAvq = 1.82E28 x (4 x 3.8E-10 x 3.8E-10) x 1.9E-5 x 1.6E-19

= 3.2E-14A (amazingly small) [2]

iv) Power =I^2 R for each conductor

so total power = 1E9 x (3.2E-14)^2 x 224 = 2.29E-16W [3]

c) Electrons undergo many collisions with atoms so mean drift velocity is very small.

EM waves travel at speed of light (ish) so v fast and very short time [2]

Not a straightforward resistivity question.

TOTAL 12

4 a) i) Ammeter in series + voltmeter in parallel with LED [1]

ii) PD across LED = 4.0v (from graph)

PD across R = 100 x 20mA = 2.0v

so Terminal PD = 6.0v [3]

b) i) E = 4.1E-19 / 1.6E-19 = 2.56eV [1]

ii) Turn on PD = 2.56v from graph

Photon is emitted when electrons have enough energy to excite atoms in LED

energy lost by electron = Energy of photon [2]

c) i) n = I / Q = 20E-3/1.6E-19 = 1.25E17 [2]

ii) Total = energy of 1 photon x no of photons /s (possible ecf)

= 1.25E17 x 4.1E-19 = 5.125E-2 [2]

iii) Efficiency = useful power out / power in = power out / VI

= 5.125E-2 / (4.0 x 20E-3) = 0.64 [2]

d) graph starts at 2.0ev and goes through 3.4v/20mA parallel to first time [2]

Not too bad if you know what you're doing but tricky all the same

TOTAL 15

5 a) yes coherent -constant phase difference

b) Superposition principle : At min add displacements together.

Min displacement = 4.0 - 2.0 = 2.0 so not zero

Waves have different amplitudes so cant cancel perfectly [2]

c) i) 90 ii) 135 [2]

d) i ) T = 0.80ms so f = 1250Hz

v = f x lambda so lambda = 340/1250 = 0.272m [3]

ii) lambda = ax/D so D = 0.40 x 2.4 x 2 / 0.272 (need max to max distance)

= 7.1m [3]

e) i) Intensity = power per unit area [1]

ii) Min amp = 2 units

Max amp = 6 units

Intensity is prop to amp squared

so max intensity = 9 x min intensity = 9 x 4.0E-6

= 3.6E-5 [3

Finishes with a real tricky bit.

TOTAL 15

Q6 nearest thing to straightforward!

a) Travel at speed of light

through a vacuum [2]

b) i) plane polarised means oscillations are confined to a single plane [2]

ii) Polariser 1 only allows vert components through and absorbs horiz

Polariser 2 only allows horizontal components through and absorbs vert

so all components absorbed and dont see anything [2]

iii) Yes - will see something.

Easiest to draw diagrams showing components - I demo this with microwaves each year.

Hopefully my students remembered it! Difficult if youve never seen it before [3]

TOTAL 9

Q7 a) Standard stuff . Waves reflect from walls.

Interference / superposition.

Cancellation - . nodes. Reinforcement -> antinodes etc [3]

b) Antinodes ; particles get hot because oscillating with max amplitudes at these points [2]

c) Some confusion about scales

Antinode to antinode = 6cm = lambda /2

so lambda = 12cm

so v = f x lambda = 2.5E9 x12E-2 = 3.0E8 as expected [4]

TOTAL 9

Not too bad!

Q8 a) i) wfe = minimum energy needed to release an electron from surface of metal [2]

ii) wfe = energy of photon that just releases electron = h x threshold frequency [2]

iii) wfe = hc/ lambda = 6.63E-34 x 3.0E8 / 55E-9 = 3.6E-19 [2]

b) i) Energy of photon = hc/lambda = 4.5E-19

so max KE = 4.5E-19 - 3.6E-19 = 9.0E-20J

so KE = 1/2 mv^2 so v = sqrt( 2 x 9.0E-20/9.11E-31 = 4.45E5 [3]

ii) lambda = h /mv = 1.63E-9m [2]

c) i) Longer wavelength so smaller energy gap so 3 -> 2 [1]

ii) 3rd transition means 3 -> 1

Calculate energies of2->1 and 3->2 , add then use E=hf

or add frequencies etc

Ans = 252nm [3]

Tough ending to a tough paper

Grade boundaries??

Last year it was 69 for an A and 39 for an E

They are going to come down from that

Could be 65 and 35.

Might be even lower

Best guess?

A 65

B 58

C 50

D 43

E 35

Good Luck.

Everyone is going to find this paper hard.

(Edited for various typos)

Usual disclaimers.

These are just my answers and are in no sense 'official'.

They may contain errors / typos.

I think this paper was one the hardest I've seen for a long time. Lots of tricky bits.

Expecting low grade boundaries.

Q1 a) i) Not a straight line through origin

so I is not prop to V

so doesn't obey Ohm's law [1]

ii) Line is approx straight so R is approx constant

R = V / I = 0.5v / 50mA = 10ohms [2]

iii) At 6v filament gets hot ( more power dissipated)

Atoms oscillate with bigger amplitude so more collisions with electrons so more energy converted to heat

so resistance increases [2]

b) In the daytime Rldr = 1ohm

so Rcombination is approx 1 ohm

so Pd divides 25:1 with resistor getting 25/26 of 12v and LDR getting 1/26 of 12v

PD across light bulb = PD across LDR is small and doesnt light

At night Rldr = 1000ohm

so R combination = R light bulb approx = 25ohm

so light bulb gets 1/2 of 12v = 6v

so lights up. [5]

Tricky potential divider with that parallel combination.

TOTAL 10

Q2 a) Emf = J/C; resistance = V/A; energy = VC; charge = As [2]

b) i) PD is the energy converted per unit charge from electrical to other forms of energy in a load. [2]

ii) Internal resistance = resistance between the terminals of the cell. [1]

iii) Same PD (in parallel) I = V/R so if half the resistance in branch (6ohm and 12 ohm)) then it

must have twice the current [2]

iv) PD across top branch = I x total R = 0.16x6 = 0.96v

so 0.96v across bottom branch

Potential divider so PD across 6ohm = 6/12 x 0.96 = 0.48v

[A lot of work for 2 marks]

v) PD splits 3:3 in top branch and 6:6 in bottom branch

same potential at each point so no difference [2]

vi) Current in bottom branch = 0.08A (half the other one)

so current through cell = 0.16 + 0.08 = 0.24A

V = E - Ir so 0.96 = 1.2 - 0.24 r

so r= 1.0ohm (possible -1 for sf of 1ohm answer) [4]

A tough question after the easy definitions

TOTAL 15

Q3 a) resistance = PD / Current [1]

b) i) strange setup which will throw many

R = rho x L / A = 1.7E-8 x 20 x 3.8E-10 / (4 x 3.8E-10) ^2

= 224 ohm [3]

ii) no density of electrons = no of atoms per unit vol

= 4 x 20 x 1 / ( 4d x d x 20 d) = 1/ 3.8E-10^3 = 1.82E28 /m3 [1]

Hard work for 1 mark

iii) I = nAvq = 1.82E28 x (4 x 3.8E-10 x 3.8E-10) x 1.9E-5 x 1.6E-19

= 3.2E-14A (amazingly small) [2]

iv) Power =I^2 R for each conductor

so total power = 1E9 x (3.2E-14)^2 x 224 = 2.29E-16W [3]

c) Electrons undergo many collisions with atoms so mean drift velocity is very small.

EM waves travel at speed of light (ish) so v fast and very short time [2]

Not a straightforward resistivity question.

TOTAL 12

4 a) i) Ammeter in series + voltmeter in parallel with LED [1]

ii) PD across LED = 4.0v (from graph)

PD across R = 100 x 20mA = 2.0v

so Terminal PD = 6.0v [3]

b) i) E = 4.1E-19 / 1.6E-19 = 2.56eV [1]

ii) Turn on PD = 2.56v from graph

Photon is emitted when electrons have enough energy to excite atoms in LED

energy lost by electron = Energy of photon [2]

c) i) n = I / Q = 20E-3/1.6E-19 = 1.25E17 [2]

ii) Total = energy of 1 photon x no of photons /s (possible ecf)

= 1.25E17 x 4.1E-19 = 5.125E-2 [2]

iii) Efficiency = useful power out / power in = power out / VI

= 5.125E-2 / (4.0 x 20E-3) = 0.64 [2]

d) graph starts at 2.0ev and goes through 3.4v/20mA parallel to first time [2]

Not too bad if you know what you're doing but tricky all the same

TOTAL 15

5 a) yes coherent -constant phase difference

b) Superposition principle : At min add displacements together.

Min displacement = 4.0 - 2.0 = 2.0 so not zero

Waves have different amplitudes so cant cancel perfectly [2]

c) i) 90 ii) 135 [2]

d) i ) T = 0.80ms so f = 1250Hz

v = f x lambda so lambda = 340/1250 = 0.272m [3]

ii) lambda = ax/D so D = 0.40 x 2.4 x 2 / 0.272 (need max to max distance)

= 7.1m [3]

e) i) Intensity = power per unit area [1]

ii) Min amp = 2 units

Max amp = 6 units

Intensity is prop to amp squared

so max intensity = 9 x min intensity = 9 x 4.0E-6

= 3.6E-5 [3

Finishes with a real tricky bit.

TOTAL 15

Q6 nearest thing to straightforward!

a) Travel at speed of light

through a vacuum [2]

b) i) plane polarised means oscillations are confined to a single plane [2]

ii) Polariser 1 only allows vert components through and absorbs horiz

Polariser 2 only allows horizontal components through and absorbs vert

so all components absorbed and dont see anything [2]

iii) Yes - will see something.

Easiest to draw diagrams showing components - I demo this with microwaves each year.

Hopefully my students remembered it! Difficult if youve never seen it before [3]

TOTAL 9

Q7 a) Standard stuff . Waves reflect from walls.

Interference / superposition.

Cancellation - . nodes. Reinforcement -> antinodes etc [3]

b) Antinodes ; particles get hot because oscillating with max amplitudes at these points [2]

c) Some confusion about scales

Antinode to antinode = 6cm = lambda /2

so lambda = 12cm

so v = f x lambda = 2.5E9 x12E-2 = 3.0E8 as expected [4]

TOTAL 9

Not too bad!

Q8 a) i) wfe = minimum energy needed to release an electron from surface of metal [2]

ii) wfe = energy of photon that just releases electron = h x threshold frequency [2]

iii) wfe = hc/ lambda = 6.63E-34 x 3.0E8 / 55E-9 = 3.6E-19 [2]

b) i) Energy of photon = hc/lambda = 4.5E-19

so max KE = 4.5E-19 - 3.6E-19 = 9.0E-20J

so KE = 1/2 mv^2 so v = sqrt( 2 x 9.0E-20/9.11E-31 = 4.45E5 [3]

ii) lambda = h /mv = 1.63E-9m [2]

c) i) Longer wavelength so smaller energy gap so 3 -> 2 [1]

ii) 3rd transition means 3 -> 1

Calculate energies of2->1 and 3->2 , add then use E=hf

or add frequencies etc

Ans = 252nm [3]

Tough ending to a tough paper

Grade boundaries??

Last year it was 69 for an A and 39 for an E

They are going to come down from that

Could be 65 and 35.

Might be even lower

Best guess?

A 65

B 58

C 50

D 43

E 35

Good Luck.

Everyone is going to find this paper hard.

(Edited for various typos)

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#2

(Original post by

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**teachercol**).

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#4

This might be wrong, but for 3.b.iv shouldn't you x I by 10^9 and then square that value for V=I^2R (and x R by 10^9 too as you have)? Getting approx. 230W.

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#5

In the I = neva

I used n = 2 x 10^28 because my own value for n wasn't right (calc error, i did 80/vol)

Will I get the marks?

Because I wrote

If n = 2 x 10^28 m^-3 then carried on calculating? And had 5.5 I think instead of 5.2

I used n = 2 x 10^28 because my own value for n wasn't right (calc error, i did 80/vol)

Will I get the marks?

Because I wrote

If n = 2 x 10^28 m^-3 then carried on calculating? And had 5.5 I think instead of 5.2

0

reply

(Original post by

This might be wrong, but for 3.b.iv shouldn't you x I by 10^9 and then square that value for V=I^2R (and x R by 10^9 too as you have)? Getting approx. 230W.

**Elcor**)This might be wrong, but for 3.b.iv shouldn't you x I by 10^9 and then square that value for V=I^2R (and x R by 10^9 too as you have)? Getting approx. 230W.

0

reply

(Original post by

In the I = neva

I used n = 2 x 10^28 because my own value for n wasn't right (calc error, i did 80/vol)

Will I get the marks?

Because I wrote

If n = 2 x 10^28 m^-3 then carried on calculating? And had 5.5 I think instead of 5.2

**L'Evil Fish**)In the I = neva

I used n = 2 x 10^28 because my own value for n wasn't right (calc error, i did 80/vol)

Will I get the marks?

Because I wrote

If n = 2 x 10^28 m^-3 then carried on calculating? And had 5.5 I think instead of 5.2

0

reply

Report

#9

(Original post by

OCR Physics A G482 EWP 9th June 2014

Usual disclaimers.

These are just my answers and are in no sense 'official'.

They may contain errors / typos.

I think this paper was one the hardest I've seen for a long time. Lots of tricky bits.

Expecting low grade boundaries.

Q1 a) i) Not a straight line through origin

so I is not prop to V

so doesn't obey Ohm's law [2]

ii) Line is approx straight so R is approx constant

R = V / I = 0.5v / 50mA = 10ohms [2]

iii) At 6v filament gets hot ( more power dissipated)

Atoms oscillate with bigger amplitude so more collisions with electrons so more energy converted to heat

so resistance increases [2]

b) In the daytime Rldr = 1ohm

so Rcombination is approx 1 ohm

so Pd divides 25:1 with resistor getting 25/26 of 12v and LDR getting 1/26 of 12v

PD across light bulb = PD across LDR is small and doesnt light

At night Rldr = 1000ohm

so R combination = R light bulb approx = 25ohm

so light bulb gets 1/2 of 12v = 6v

so lights up. [5]

Tricky potential divider with that parallel combination.

TOTAL 10

Q2 a) Emf = J/C; resistance = V/A; energy = VC; charge = As [2]

b) i) PD is the energy converted per unit charge from electrical to other forms of energy in a load. [2]

ii) Internal resistance = resistance between the terminals of the cell. [1]

iii) Same PD (in parallel) I = V/R so if half the resistance in branch (6ohm and 12 ohm)) then it

must have twice the current [2]

iv) PD across top branch = I x total R = 0.16x6 = 0.96v

so 0.96v across bottom branch

Potential divider so PD across 6ohm = 6/12 x 0.96 = 0.48v

[A lot of work for 2 marks]

v) PD splits 3:3 in top branch and 6:6 in bottom branch

same potential at each point so no difference [2]

vi) Current in bottom branch = 0.08A (half the other one)

so current through cell = 0.16 + 0.08 = 0.24A

V = E - Ir so 0.96 = 1.2 - 0.24 r

so r= 1.0ohm (possible -1 for sf of 1ohm answer) [4]

A tough question after the easy definitions

TOTAL 15

Q3 a) resistance = PD / Current [1]

b) i) strange setup which will throw many

R = rho x L / A = 1.7E-8 x 20 x 3.8E-10 / (4 x 3.8E-10) ^2

= 224 ohm [3]

ii) no density of electrons = no of atoms per unit vol

= 4 x 20 x 1 / ( 4d x d x 20 d) = 1/ 3.8E-10^3 = 1.82E28 /m3 [1]

Hard work for 1 mark

iii) I = nAvq = 1.82E28 x (4 x 3.8E-10 x 3.8E-10) x 1.9E-5 x 1.6E-19

= 3.2E-14A (amazingly small) [2]

iv) Power =I^2 R for each conductor

so total power = 1E9 x (3.2E-14)^2 x 224 = 2.29E16W [3]

c) Electrons undergo many collisions with atoms so mean drift velocity is very small.

EM waves travel at speed of light (ish) so v fast and very short time [2]

Not a straightforward resistivity question.

TOTAL 12

4 a) i) Ammeter in series + voltmeter in parallel with LED [1]

ii) PD across LED = 4.0v (from graph)

PD across R = 100 x 20mA = 2.0v

so Terminal PD = 6.0v [3]

b) i) E = 4.1E-19 / 1.6E-19 = 2.56eV [1]

ii) Turn on PD = 2.56v from graph

Photon is emitted when electrons have enough energy to excite atoms in LED

energy lost by electron = Energy of photon [2]

c) i) n = I / Q = 20E-3/1.6E-19 = 1.25E17 [2]

ii) Total = energy of 1 photon x no of photons /s (possible ecf)

= 1.25E17 x 4.1E-19 = 5.125E-2 [2]

iii) Efficiency = useful power out / power in = power out / VI

= 5.125E-2 / (4.0 x 20E-3) = 0.64 [2]

d) graph starts at 2.0ev and goes through 3.4v/20mA parallel to first time [2]

Not too bad if you know what you're doing but tricky all the same

TOTAL 15

5 a) yes coherent -constant phase difference

b) Superposition principle : At min add displacements together.

Min displacement = 4.0 - 2.0 = 2.0 so not zero

Waves have different amplitudes so cant cancel perfectly [2]

c) i) 90 ii) 135 [2]

d) i ) T = 0.80ms so f = 1250Hz

v = f x lambda so lambda = 340/1250 = 0.272m [3]

ii) lambda = ax/D so D = 0.40 x 2.4 x 2 / 0.272 (need max to max distance)

= 7.1m [3]

e) i) Intensity = power per unit area [1]

ii) Min amp = 2 units

Max amp = 6 units

Intensity is prop to amp squared

so max intensity = 9 x min intensity = 9 x 4.0E-6

= 3.6E-5 [3

Finishes with a real tricky bit.

TOTAL 15

Q6 nearest thing to straightforward!

a) Travel at speed of light

through a vacuum [2]

b) i) plane polarised means oscillations are confined to a single plane [2]

ii) Polariser 1 only allows vert components through and absorbs horiz

Polariser 2 only allows horizontal components through and absorbs vert

so all components absorbed and dont see anything [2]

iii) Yes - will see something.

Easiest to draw diagrams showing components - I demo this with microwaves each year.

Hopefully my students remembered it! Difficult if youve never seen it before [3]

TOTAL 9

Q7 a) Standard stuff . Waves reflect from walls.

Interference / superposition.

Cancellation - . nodes. Reinforcement -> antinodes etc [3]

b) Antinodes ; particles get hot because oscillating with max amplitudes at these points [2]

c) Some confusion about scales

Antinode to antinode = 6cm = lambda /2

so lambda = 12cm

so v = f x lambda = 2.5E9 x12E-2 = 3.0E8 as expected [4]

TOTAL 9

Not too bad!

Q8 a) i) wfe = minimum energy needed to release an electron from surface of metal [2]

ii) wfe = energy of photon that just releases electron = h x threshold frequency [2]

iii) wfe = hc/ lambda = 6.63E-34 x 3.0E8 / 55E-9 = 3.6E-19

b) i) Energy of photon = hc/lambda = 4.5E-19

so max KE = 4.5E-19 - 3.6E-19 = 9.0E-20J

so KE = 1/2 mv^2 so v = sqrt( 2 x 9.0E-20/9.11E-31 = 4.45E5 [3]

ii) lambda = h /mv = 1.63E-9m [2]

c) i) Longer wavelength so smaller energy gap so 3 -> 2 [1]

ii) 3rd transition means 3 -> 1

Calculate energies of2->1 and 3->2 , add then use E=hf

or add frequencies etc

Ans = 252nm [3]

Tough ending to a tough paper

Grade boundaries??

Last year it was 69 for an A and 39 for an E

They are going to come down from that

Could be 65 and 35.

Might be even lower

Best guess?

A 65

B 58

C 50

D 43

E 35

Good Luck.

Everyone is going to find this paper hard.

**teachercol**)OCR Physics A G482 EWP 9th June 2014

Usual disclaimers.

These are just my answers and are in no sense 'official'.

They may contain errors / typos.

I think this paper was one the hardest I've seen for a long time. Lots of tricky bits.

Expecting low grade boundaries.

Q1 a) i) Not a straight line through origin

so I is not prop to V

so doesn't obey Ohm's law [2]

ii) Line is approx straight so R is approx constant

R = V / I = 0.5v / 50mA = 10ohms [2]

iii) At 6v filament gets hot ( more power dissipated)

Atoms oscillate with bigger amplitude so more collisions with electrons so more energy converted to heat

so resistance increases [2]

b) In the daytime Rldr = 1ohm

so Rcombination is approx 1 ohm

so Pd divides 25:1 with resistor getting 25/26 of 12v and LDR getting 1/26 of 12v

PD across light bulb = PD across LDR is small and doesnt light

At night Rldr = 1000ohm

so R combination = R light bulb approx = 25ohm

so light bulb gets 1/2 of 12v = 6v

so lights up. [5]

Tricky potential divider with that parallel combination.

TOTAL 10

Q2 a) Emf = J/C; resistance = V/A; energy = VC; charge = As [2]

b) i) PD is the energy converted per unit charge from electrical to other forms of energy in a load. [2]

ii) Internal resistance = resistance between the terminals of the cell. [1]

iii) Same PD (in parallel) I = V/R so if half the resistance in branch (6ohm and 12 ohm)) then it

must have twice the current [2]

iv) PD across top branch = I x total R = 0.16x6 = 0.96v

so 0.96v across bottom branch

Potential divider so PD across 6ohm = 6/12 x 0.96 = 0.48v

[A lot of work for 2 marks]

v) PD splits 3:3 in top branch and 6:6 in bottom branch

same potential at each point so no difference [2]

vi) Current in bottom branch = 0.08A (half the other one)

so current through cell = 0.16 + 0.08 = 0.24A

V = E - Ir so 0.96 = 1.2 - 0.24 r

so r= 1.0ohm (possible -1 for sf of 1ohm answer) [4]

A tough question after the easy definitions

TOTAL 15

Q3 a) resistance = PD / Current [1]

b) i) strange setup which will throw many

R = rho x L / A = 1.7E-8 x 20 x 3.8E-10 / (4 x 3.8E-10) ^2

= 224 ohm [3]

ii) no density of electrons = no of atoms per unit vol

= 4 x 20 x 1 / ( 4d x d x 20 d) = 1/ 3.8E-10^3 = 1.82E28 /m3 [1]

Hard work for 1 mark

iii) I = nAvq = 1.82E28 x (4 x 3.8E-10 x 3.8E-10) x 1.9E-5 x 1.6E-19

= 3.2E-14A (amazingly small) [2]

iv) Power =I^2 R for each conductor

so total power = 1E9 x (3.2E-14)^2 x 224 = 2.29E16W [3]

c) Electrons undergo many collisions with atoms so mean drift velocity is very small.

EM waves travel at speed of light (ish) so v fast and very short time [2]

Not a straightforward resistivity question.

TOTAL 12

4 a) i) Ammeter in series + voltmeter in parallel with LED [1]

ii) PD across LED = 4.0v (from graph)

PD across R = 100 x 20mA = 2.0v

so Terminal PD = 6.0v [3]

b) i) E = 4.1E-19 / 1.6E-19 = 2.56eV [1]

ii) Turn on PD = 2.56v from graph

Photon is emitted when electrons have enough energy to excite atoms in LED

energy lost by electron = Energy of photon [2]

c) i) n = I / Q = 20E-3/1.6E-19 = 1.25E17 [2]

ii) Total = energy of 1 photon x no of photons /s (possible ecf)

= 1.25E17 x 4.1E-19 = 5.125E-2 [2]

iii) Efficiency = useful power out / power in = power out / VI

= 5.125E-2 / (4.0 x 20E-3) = 0.64 [2]

d) graph starts at 2.0ev and goes through 3.4v/20mA parallel to first time [2]

Not too bad if you know what you're doing but tricky all the same

TOTAL 15

5 a) yes coherent -constant phase difference

b) Superposition principle : At min add displacements together.

Min displacement = 4.0 - 2.0 = 2.0 so not zero

Waves have different amplitudes so cant cancel perfectly [2]

c) i) 90 ii) 135 [2]

d) i ) T = 0.80ms so f = 1250Hz

v = f x lambda so lambda = 340/1250 = 0.272m [3]

ii) lambda = ax/D so D = 0.40 x 2.4 x 2 / 0.272 (need max to max distance)

= 7.1m [3]

e) i) Intensity = power per unit area [1]

ii) Min amp = 2 units

Max amp = 6 units

Intensity is prop to amp squared

so max intensity = 9 x min intensity = 9 x 4.0E-6

= 3.6E-5 [3

Finishes with a real tricky bit.

TOTAL 15

Q6 nearest thing to straightforward!

a) Travel at speed of light

through a vacuum [2]

b) i) plane polarised means oscillations are confined to a single plane [2]

ii) Polariser 1 only allows vert components through and absorbs horiz

Polariser 2 only allows horizontal components through and absorbs vert

so all components absorbed and dont see anything [2]

iii) Yes - will see something.

Easiest to draw diagrams showing components - I demo this with microwaves each year.

Hopefully my students remembered it! Difficult if youve never seen it before [3]

TOTAL 9

Q7 a) Standard stuff . Waves reflect from walls.

Interference / superposition.

Cancellation - . nodes. Reinforcement -> antinodes etc [3]

b) Antinodes ; particles get hot because oscillating with max amplitudes at these points [2]

c) Some confusion about scales

Antinode to antinode = 6cm = lambda /2

so lambda = 12cm

so v = f x lambda = 2.5E9 x12E-2 = 3.0E8 as expected [4]

TOTAL 9

Not too bad!

Q8 a) i) wfe = minimum energy needed to release an electron from surface of metal [2]

ii) wfe = energy of photon that just releases electron = h x threshold frequency [2]

iii) wfe = hc/ lambda = 6.63E-34 x 3.0E8 / 55E-9 = 3.6E-19

b) i) Energy of photon = hc/lambda = 4.5E-19

so max KE = 4.5E-19 - 3.6E-19 = 9.0E-20J

so KE = 1/2 mv^2 so v = sqrt( 2 x 9.0E-20/9.11E-31 = 4.45E5 [3]

ii) lambda = h /mv = 1.63E-9m [2]

c) i) Longer wavelength so smaller energy gap so 3 -> 2 [1]

ii) 3rd transition means 3 -> 1

Calculate energies of2->1 and 3->2 , add then use E=hf

or add frequencies etc

Ans = 252nm [3]

Tough ending to a tough paper

Grade boundaries??

Last year it was 69 for an A and 39 for an E

They are going to come down from that

Could be 65 and 35.

Might be even lower

Best guess?

A 65

B 58

C 50

D 43

E 35

Good Luck.

Everyone is going to find this paper hard.

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(Original post by

For 7c, I converted the frequency to a period and did wavelength/ period, 12x10^-2/whatever it was and got 3x10^8. No idea why I did this but it's the same thing so will I get the marks?

**Red Fox**)For 7c, I converted the frequency to a period and did wavelength/ period, 12x10^-2/whatever it was and got 3x10^8. No idea why I did this but it's the same thing so will I get the marks?

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#11

(Original post by

I'd expect ecf there . so yes you should get the marks

**teachercol**)I'd expect ecf there . so yes you should get the marks

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#12

(Original post by

I don't think so.

**teachercol**)I don't think so.

My logic was to look at the chip as a whole, so the current will be 10^9 x larger and so will R. Then to P=I^2R on the chip as a whole, getting ~230W. Is that wrong?

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(Original post by

Was question one 10 or 11 marks?

**SH0405**)Was question one 10 or 11 marks?

Typo in a) its only 1 mark. I'll edit it.

Thanks

Edited

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#15

(Original post by

Its 10 marks.

Typo in a) its only 1 mark. I'll edit it.

Thanks

It wont let me .....bah

**teachercol**)Its 10 marks.

Typo in a) its only 1 mark. I'll edit it.

Thanks

It wont let me .....bah

Ok - also for question 6.iii), I believe I explained it quite well. I brought in Malus' Law - is this right? However I forgot to square cos(theta) when explaining the Law. Assuming I explained the polarisation theory correctly, how many marks would you say this is?

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(Original post by

But if you think about it, 2.3E16 W, the value you got, is a massive power for a small chip used in a computer.

My logic was to look at the chip as a whole, so the current will be 10^9 x larger and so will R. Then to P=I^2R on the chip as a whole, getting ~230W. Is that wrong?

**Elcor**)But if you think about it, 2.3E16 W, the value you got, is a massive power for a small chip used in a computer.

My logic was to look at the chip as a whole, so the current will be 10^9 x larger and so will R. Then to P=I^2R on the chip as a whole, getting ~230W. Is that wrong?

The current wont be 1E9 times higher.

Youre told each chip has the same current (3.2E-14A) calculated in iii

You just multiply the number of chips by the power for one chip.

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(Original post by

Ok - also for question 6.iii), I believe I explained it quite well. I brought in Malus' Law - is this right? However I forgot to square cos(theta) when explaining the Law. Assuming I explained the polarisation theory correctly, how many marks would you say this is?

**SH0405**)Ok - also for question 6.iii), I believe I explained it quite well. I brought in Malus' Law - is this right? However I forgot to square cos(theta) when explaining the Law. Assuming I explained the polarisation theory correctly, how many marks would you say this is?

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