Seriously has no one done an unofficial markscheme??
also what did everyone get for the very last question? I think i did 0.8^2 / 0.6^2 *30mm = 53.3
OCR Physics A G482, Electrons, Waves and Photons, 25th May 2012


(Original post by Sarabande)
How does it make sense in regards to the question that was asked. It was asking which law did you take into account in your proof of the 0.2A resistor. You didn't account for the 1st law ONCE if you proved it correctly whether through use of ratios between resistance and current (V=IR) or flat out using V=IR 
(Original post by HarryW95)
You're making this so much more difficult than it needs to be.. Because you seem to be some what incapable to interpreting a graph, I've got the values for you.
http://gyazo.com/eafb7b433afb3946ae23333e57aa99f6
[R=V/I]
At 1.8 V = 1.8/9x10^3 = 200 ohms
At 2.0 V = 2.0/38x10^3 = 52.63 ohms
Hence the resistance is decreasing. Are we done here..?
If I'd have done calculations, I'd have taken at least 3 points, found the dY/dX of the first two, and then dY/dX of the last two, and compared to see if the gradient, and thus R, is constant.
If the line doesn't go through the origin then 2 different Y/X values aren't going to be the same, even for a simple linear equation such as 2x 6. So we can't do (I/V)^1 here to find resistance (in two places).
To be honest, the only reason I went with R is constant is because it didn't state to use calculations, it stated to use features of the graph, or 'AW' as the mark scheme would say.
How come you just did I/V etc? I can't see what you're getting at by that  of course they'll be different, no? 
Yeah, people who didn't just know that the straight line that didn't pass through the origin meant a decreasing resistance, how did you know to sub in the values in R = V/I to check for a decreasing resistance because it didn't state to?

(Original post by jbelks)
Yeah, at first i thought i did something wrong as the resistance was really high.
How about the question about the 60W filament? Cross sectional area increases as resistance decreases, due to R being inversely proportional to A. Then i backed it up with the idea that current was greater with 60W hence R=V/I, R would be smaller 
(Original post by ahct_952)
yeah, that's what i put too, but some people at my school said they put 15mm because they thought the question meant the distance between two nodes/two antinodes was 7.5mm, so now i'm not sure... i think the wording of the question was 'the distance between adjacent maxima and minima is 7.5mm', which in my opinion does not make clear what they mean 
(Original post by Z REFAN Z)
yeah the wording of the question isn't very clear. 
Answers by teachercol are up folks: http://www.thestudentroom.co.uk/show....php?t=2014571

(Original post by geditor)
Yeah, people who didn't just know that the straight line that didn't pass through the origin meant a decreasing resistance, how did you know to sub in the values in R = V/I to check for a decreasing resistance because it didn't state to? 
these are teachercol's answers.
Usual disclaimer. These are just my answers and are in no sense official.
They may contain erros and typos.
Q1 a) Power = rate of doing work or Work done per unit time (1)
b) i) P=VI so I = 40/230 = 0.17(4) A (2)
ii) RV/I (or use P = I^2R) R = 230/0.174 = 1320 ohm (2)
c) R = rho L/A so L = 1320 x 3.0E8/7.0E5 = 0.567m (3) ecf
d) If same PD bigger power must have more current (P=VI)
so R must be smaller so A must be bigger so must be thicker (3)
e) i) Q=It = 0.174 x 8 x 3600 = 5011C (2) ecf
ii) 1) kW.h = ENERGY converted by a device of power 1kW operating
for a time of 1h (1)
2) Energy = 40/1000 x 8 = 0.32kWh
so cost = 0.32 x 22 = 7(.04)p (2)
Total 15
Q2 a) i) 1) R is infinite below 1.4V (1)
2) When V=1.8v, I = 10mA so R=1.8/10E3 = 180ohm (2)
ii) R gets smaller. Current is increasing faster than PD.
or just calculate R at different points. (3)
NB Ohm's law doesnt apply here graph not a striaght line through the origin
R is NOT the gradient. R=V/I not dV/dI
b) Draw LED (right way round ) with ammeter/ voltmeter and resistor in right places.(4)
c) LED used for torch bulbs / car headlights.
More efficient becasue they produce less heat. Lower running costs. Longer lifetime. (2)
Total 12
Q3 a) When temp rises, Rtherm falls, Total R falls, cureent increases. (3)
b) Either use potential divider or use Total R and work out current then PD.
Rtherm = 40ohm so ratio of R is 200:40 so PD must split in same ratio 5:1
so PD=5v (4)
c) i) Draw LDR symbol (1)
ii) When light intensity increases, R falls so ratio Rldr:R200 must drop
so PD across 200ohm resistor must increase.
or total R drops so current increases so PD across 200 increases. (2)
Total 10
Q4 a) PD across 6ohm = IR = 0.30A x 6.0ohm = 1.2v
In parallel so same PD so I = V/R = 1.2/6.0 = 0.20A
b) i) Kirchoff's 1st law (1)
ii) I = I1 + I2 = 0.30 + 0.20 = 0.50A (1)
c) 1/Rp = 1/R1 + 1/R2 = 1/6 + 1/4 = 5/12
Rp = 12/5 = 2.4ohm
Total R = 2.4+5.6 = 8.0 ohm (3)
d) i) Emf = energy converted per unit charge
from one form of energy to electrical energy in a source (2)
ii) Terminal PD = PD across combination = 0.50 x 8.0 = 4.0v (1)
(or add up Pds)
iii) V = E  Ir
4.0 = 5.0  0.50r so r = 2.0 ohm (2)
Total 12
Q5 a) e has charge / photon doesnt
e has mass / photon doesnt
photn travels at c / e cant (2)
b) i) Gain in energy = QV = 1.6E19 v 5000 = 8.0E16J (2)
ii) Energy = 1/2 m v^2
8.0E16 = 1/2 x 9,1E31 x v^2 so v = 4.2E7 ms1 (3) ecf
c) i) Wavelength of probability wave associated with electron (1)
(not sure what they expect here. )
d) E = hf
8.0E16 = 6.6E34 x f so f = 1.21E18 Hz
lambda = c/f = 3.0E8/1.21E18 = 2.48E10m (3)
e) i) Photoelectric effect (1)
ii) max KE = hf  phi
= 9.0E19  7.2E19 = 1.8E19
iii) phi is min energy needed to release an electron from surface.
deeper electrons need more energy
so have less KE. (2)
Total 19
Q6 a) i) Displacment = distance moved from equilibrium in a specified direction
Amplitude = max displacment (2)
ii) Frequency = no of complete oscillation per unit time
Phase difference = measure of how much out of step two
oscialltions are measured in fractions of a cycle
or as ana angle 0360 or 0  2pi (2)
b) Diagram is essentially mirror imaged. Wave travels 0.25m every 0.5s (4)
Total 8
Q7 a) i) Photon is emitted when atom moves from higher to lower energy level
Energy of photon = hf = hc/lambda= difference in energy of two levels.
so One wavelength corresponds to one transition between higher / lower energy level
three possible transitions (32, 31 and 21) so three different wavelengths (3)
ii) 1) f = v/lambda = 3.0E8/6.56E7 = 4.57E14Hz
E = hf = 6.6E34 x 4.57E14 = 3.0E19J (2)
2) Photon energy Must be difference btween energy of levels
so must be 32 (1)
b) i) 1) lambda = d sin theta so d = 6.56E7 / sin 11.4 = 3.32E6
2) N = 1/d = 301306 lines per m
= 301 lines per mm (1)
ii) Blue has smaller lambda
sin theta = lambda / d d is same so theta is smaller (1)
Total 11
Q8 a) same speed / travel through vacuum / E and B fields oscillate (2)
b) uwave = 1E4 / UV = 1E8 / gamma = 1E12 / IR = 1E6 (2)
c) i) Receiver reflects incident wave back at 90 but some is absorbed (freq stays same)
so reflected wave has lower amplitude and same freq than incident wave (2)
ii) Waves superpose. R detects resultant amplitude
At some points waves are in phase / reinforce and give maxima
At other points waves are 180 out / cancel and we get minima (3)
iii) max  min = 7.5mm
so lambda = 4 x 7.5 = 30 mm (1)
iV) Max amp = 0.8a + 0.6a = 1.4a
Min amp = 0.8a  0.6a = 0.2a
I is prop to square of amp
so ratio = (1.4a/).2a) ^2 = 49 (3)
Total 13
I thought this was quite an easy paper but the feedback I've had was mixed.
Some people really liked it, others hated it.
There wasnt much in the way of descriptive stuff, a lot of calcualtion.
It probably suits mathematically inclined students.
I still think its easy so grade boundaries may not be as low as they have been.
No  I cant tell you how many UMS you'll get. Dont ask. 
told you kircchhoffs 1st law

(Original post by mashmammad)
these are teachercol's answers.
Usual disclaimer. These are just my answers and are in no sense official.
They may contain erros and typos.
Q1 a) Power = rate of doing work or Work done per unit time (1)
b) i) P=VI so I = 40/230 = 0.17(4) A (2)
ii) RV/I (or use P = I^2R) R = 230/0.174 = 1320 ohm (2)
c) R = rho L/A so L = 1320 x 3.0E8/7.0E5 = 0.567m (3) ecf
d) If same PD bigger power must have more current (P=VI)
so R must be smaller so A must be bigger so must be thicker (3)
e) i) Q=It = 0.174 x 8 x 3600 = 5011C (2) ecf
ii) 1) kW.h = ENERGY converted by a device of power 1kW operating
for a time of 1h (1)
2) Energy = 40/1000 x 8 = 0.32kWh
so cost = 0.32 x 22 = 7(.04)p (2)
Total 15
Q2 a) i) 1) R is infinite below 1.4V (1)
2) When V=1.8v, I = 10mA so R=1.8/10E3 = 180ohm (2)
ii) R gets smaller. Current is increasing faster than PD.
or just calculate R at different points. (3)
NB Ohm's law doesnt apply here graph not a striaght line through the origin
R is NOT the gradient. R=V/I not dV/dI
b) Draw LED (right way round ) with ammeter/ voltmeter and resistor in right places.(4)
c) LED used for torch bulbs / car headlights.
More efficient becasue they produce less heat. Lower running costs. Longer lifetime. (2)
Total 12
Q3 a) When temp rises, Rtherm falls, Total R falls, cureent increases. (3)
b) Either use potential divider or use Total R and work out current then PD.
Rtherm = 40ohm so ratio of R is 200:40 so PD must split in same ratio 5:1
so PD=5v (4)
c) i) Draw LDR symbol (1)
ii) When light intensity increases, R falls so ratio Rldr:R200 must drop
so PD across 200ohm resistor must increase.
or total R drops so current increases so PD across 200 increases. (2)
Total 10
Q4 a) PD across 6ohm = IR = 0.30A x 6.0ohm = 1.2v
In parallel so same PD so I = V/R = 1.2/6.0 = 0.20A
b) i) Kirchoff's 1st law (1)
ii) I = I1 + I2 = 0.30 + 0.20 = 0.50A (1)
c) 1/Rp = 1/R1 + 1/R2 = 1/6 + 1/4 = 5/12
Rp = 12/5 = 2.4ohm
Total R = 2.4+5.6 = 8.0 ohm (3)
d) i) Emf = energy converted per unit charge
from one form of energy to electrical energy in a source (2)
ii) Terminal PD = PD across combination = 0.50 x 8.0 = 4.0v (1)
(or add up Pds)
iii) V = E  Ir
4.0 = 5.0  0.50r so r = 2.0 ohm (2)
Total 12
Q5 a) e has charge / photon doesnt
e has mass / photon doesnt
photn travels at c / e cant (2)
b) i) Gain in energy = QV = 1.6E19 v 5000 = 8.0E16J (2)
ii) Energy = 1/2 m v^2
8.0E16 = 1/2 x 9,1E31 x v^2 so v = 4.2E7 ms1 (3) ecf
c) i) Wavelength of probability wave associated with electron (1)
(not sure what they expect here. )
d) E = hf
8.0E16 = 6.6E34 x f so f = 1.21E18 Hz
lambda = c/f = 3.0E8/1.21E18 = 2.48E10m (3)
e) i) Photoelectric effect (1)
ii) max KE = hf  phi
= 9.0E19  7.2E19 = 1.8E19
iii) phi is min energy needed to release an electron from surface.
deeper electrons need more energy
so have less KE. (2)
Total 19
Q6 a) i) Displacment = distance moved from equilibrium in a specified direction
Amplitude = max displacment (2)
ii) Frequency = no of complete oscillation per unit time
Phase difference = measure of how much out of step two
oscialltions are measured in fractions of a cycle
or as ana angle 0360 or 0  2pi (2)
b) Diagram is essentially mirror imaged. Wave travels 0.25m every 0.5s (4)
Total 8
Q7 a) i) Photon is emitted when atom moves from higher to lower energy level
Energy of photon = hf = hc/lambda= difference in energy of two levels.
so One wavelength corresponds to one transition between higher / lower energy level
three possible transitions (32, 31 and 21) so three different wavelengths (3)
ii) 1) f = v/lambda = 3.0E8/6.56E7 = 4.57E14Hz
E = hf = 6.6E34 x 4.57E14 = 3.0E19J (2)
2) Photon energy Must be difference btween energy of levels
so must be 32 (1)
b) i) 1) lambda = d sin theta so d = 6.56E7 / sin 11.4 = 3.32E6
2) N = 1/d = 301306 lines per m
= 301 lines per mm (1)
ii) Blue has smaller lambda
sin theta = lambda / d d is same so theta is smaller (1)
Total 11
Q8 a) same speed / travel through vacuum / E and B fields oscillate (2)
b) uwave = 1E4 / UV = 1E8 / gamma = 1E12 / IR = 1E6 (2)
c) i) Receiver reflects incident wave back at 90 but some is absorbed (freq stays same)
so reflected wave has lower amplitude and same freq than incident wave (2)
ii) Waves superpose. R detects resultant amplitude
At some points waves are in phase / reinforce and give maxima
At other points waves are 180 out / cancel and we get minima (3)
iii) max  min = 7.5mm
so lambda = 4 x 7.5 = 30 mm (1)
iV) Max amp = 0.8a + 0.6a = 1.4a
Min amp = 0.8a  0.6a = 0.2a
I is prop to square of amp
so ratio = (1.4a/).2a) ^2 = 49 (3)
Total 13
I thought this was quite an easy paper but the feedback I've had was mixed.
Some people really liked it, others hated it.
There wasnt much in the way of descriptive stuff, a lot of calcualtion.
It probably suits mathematically inclined students.
I still think its easy so grade boundaries may not be as low as they have been.
No  I cant tell you how many UMS you'll get. Dont ask. 
FML

I went through that markscheme and...
I messed up on ton of it, especially the last question as I got the wrong idea and started talking about a stationary wave being set up between the transmitter and the metal sheet thing and talking about nodes and antinodes for maxima and minima... : P
I really hope that the boundary for an A will be a maximum of 70, then I'll just manage to scrap it by one or two marks. orz 
(Original post by geditor)
FML 
(Original post by Jullith)
I went through that markscheme and...
I messed up on ton of it, especially the last question as I got the wrong idea and started talking about a stationary wave being set up between the transmitter and the metal sheet thing and talking about nodes and antinodes for maxima and minima... : P
I really hope that the boundary for an A will be a maximum of 70, then I'll just manage to scrap it by one or two marks. orz 
(Original post by pixelfrag)
I'm sure you had to explain the law as well? 
In my textbook it says that when microwaves hit a metal sheet they refelct and form a standing wave :/

It's safe to say I've messed this exam up big time, I just hope around half marks will be a C... if I got a C in Mechanics in January no way will I have got one now.
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