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Higher Physics 2008 - How did you find it?

harveym

I completely messed up question24a i) i said the resistance was 3.2, will i still get marks for using it in the second and third part of the question

I didn't think the answer to a i) was to be used in part iii)? I used it, but it's given me a different answer to everyone else on here. It also throws my answer for b) out of whack too. I think that you will get the carry on marks, though.

Reply 181

harveym

I'm pretty sure i've failed now, i made so many stupid mistakes

Reply 182

Ross951

Hey everyone, how did you all find that paper?
i thought the multiple choice was really straight forward and i compared my answers with my bro and we are quite good at physics and both got exactly the same 20 multiple choice answers :

1D
2A
3B
4C
5D
6C
7D
8E
9E
10D
11A
12B
13E
14C
15A
16B
17B
18C
19C
20D

i got the right answer to that LED questions -- i knew how to do it, but me being me mis read the question and worked it out for RED and BLUE. not YELLOW and BLUE!!! will i get any credit for it?

:cool:

Reply 183

harveym

was the answer to the LED question the Blue light, i just guessed and said that because it was in the correct position

Reply 184

Meteorshower

14

Ross951

Hey everyone, how did you all find that paper?
i thought the multiple choice was really straight forward and i compared my answers with my bro and we are quite good at physics and both got exactly the same 20 multiple choice answers :

1D
2A
3B
4C
5D
6C
7D
8E
9E
10D
11A
12B
13E
14C
15A
16B
17B
18C
19C
20D

i got the right answer to that LED questions -- i knew how to do it, but me being me mis read the question and worked it out for RED and BLUE. not YELLOW and BLUE!!! will i get any credit for it?

:cool:

The multi choice is right :smile:

Unlucky about the LED question, i don't think you'll get it to be honest though because most of the time these questions require you to have "such and such will happen" before you get any marks, in this case "the yellow LED lit up" or something to that effect is likely what you need to say.

I think they have to assume you didn't know it was the yellow, even though it's obvious that's what it meant, cos if they assume you know something you didn't explicitly say they'd have to do that for everyone.

Reply 185

Meteorshower

14

harveym

was the answer to the LED question the Blue light, i just guessed and said that because it was in the correct position

No it was yellow, because the current is travelling from the 0 potential to the -2.1 potential. The yellow was the correct way for that, sorry

Reply 186

Lesjordans

I said Yellow but no explanation, but i never got a Negative voltage, whoops! Does this mean they'll take marks off me because they know ive guessed :s-smilie:

Reply 187

Meteorshower

14

They might give you half a mark or one depending on the scheme for getting the right answer but not justifying it.

Reply 188

Ross951

for the led question i did a huge calculation, then a small explanation and my choice at the end of it!!!

Reply 189

Ross951

is the diodes being forward or reversed biased got anything to do with it?

Reply 190

SheepBeast

hang on im gonna post my answers

Reply 191

SheepBeast

ok here are the MC answers:

1 D
2 A
3 B
4 C
5 D
6 C
7 D
8 E
9 E
10 D
11 A
12 B
13 E
14 C
15 A
16 B
17 B
18 C
19 C
20 D

and some section B answers (i left out the explanation questions):

21)
a) 42m
c)II) 15.9 ohms

22)
a)I) 196N
II) 44N

23)
a)I) 2.68x10^6 Pascals
b)I) 1.128kg

24)
a)I) 4 ohms
II) 0.75A
III) 2.025W

25)
b)I) 3.4V
II)2125 ohms
III) 0.15mF

26)
a) 0.48W/m^2
b)II) -2.1V
III) yellow LED is lit

27)
a)I) 16.95 degrees
II)388.2nm

28)
a) 25W/m^2
b) no because k is not constant in the equation I=k/d^2

29)
b) 3.44x10^-19
c) max kinetic energy depends on frequency not irradiance

30)
a)I) 12000 particles decay each second
II) 25 Bq
b)I) 0.03 Sv
II) 2000 times

hope that helps people :smile:

im pretty sure they're all right

Reply 192

SheepBeast

to Ross951,

yes you get the answer of the yellow one being lit because it is forward biased, while the blue is reverse biased to cannot light

Reply 193

Zooba12

Can anyone show me how they did 25 b(III) The Capacitance one please.
I think i got it correct but not sure now :confused:

Also did the capacitor take less time to charge in 25 (c) cause the resistance has lowered cause of resistors in parallel.

Reply 194

sjames06

SheepBeast

ok here are the MC answers:

30)
b)I) 0.03 Sv

Probably a typo, but shouldn't that be 0.03 µSv?

Reply 195

Zooba12

Can anyone put up images of the multiple choice questions please?

Reply 196

Xeraii

I've been pondering this since I sat the exam and it's really getting on my nerves. For the LED question I said no LEDs would light. I know this is the "wrong" answer but I think it reveals a flaw in the question to be honest. Here are my thoughts:

•

The yellow LED is reverse biased and the blue LED is forward biased.

•

Since the voltage is negative this switches it around therefore it would appear the yellow LED will light since it is now forward biased.

•

However, if you mess around with the formula E = QV you can deduce the following:

•

If the voltage is -2.1V the energy supplied to the yellow LED is E = (-1.6x10^-19 . -2.1) = 3.36x10^-19 J.

•

The LED produces yellow light, meaning a photon of yellow light is emitted. From here you can calculate the energy of a photon of yellow light using E = hf, where h is Plank's constant.

•

Now, if you use the wavelength value for yellow light given at the front of the paper, 589 nanometers, you can calculate the frequency of the photon.

•

f = velocity / wavelength = 3x10^8 / 589x10^-9 = 5.09x10^14 Hz

•

Now calculate the energy of the photon:

•

E = 6.63x10^-34 . 5.09x10^14 = 3.378x10^-19 J

It would appear that the energy supplied by the -2.1V is less than the minimum energy required to release a photon of yellow light?

To quickly recap:

-2.1V gives you 3.36x10^-19 J of energy.
A yellow photon, following the values given in the test paper has 3.378x10^-19 J of energy.
How can this -2.1V release a photon of yellow light?

I'm sure I'm missing something here. Granted, the difference in energy is miniscule, and yellow light has wavelengths above and below 589nm. But surely the questions asked should be consistent with the wavelength values given at the front of the paper?

Am I missing something? I'm sure there's some major flaw in my logic here. :p:

Reply 197

Meteorshower

14

You're absolutely correct there as far as i'm aware, it gives a wavelength of 592, but because the colour yellow is not only at 589 nm (that is only the spectral line of sodium in the data booklet, which happens to be yellow) then it's acceptable to say it lights up.

I like your style of thinking though, you'll go far with that

Reply 198

Zooba12

For the work function question did you use Ek=hf-hfo to get the lowest frequency it emits at then plug that answer into E=hf to get the work function? Thats what i did not sure if it was correct or not.

Reply 199

Meteorshower

14

Zooba12

For the work function question did you use Ek=hf-hfo to get the lowest frequency it emits at then plug that answer into E=hf to get the work function? Thats what i did not sure if it was correct or not.

That seems right to me

Higher Physics 2008 - How did you find it?

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