AQA Physics A - PHYA5 (18/06/12) - Exam thread

Yeah mate



Those grade boundaries won't be repeated, if an A is 40/1 on this paper not 44 like it was on that, you're getting 1 or 2 marks above it, there fore around 50.

ASTRO MS:
1ai) http://mcat-review.org/convex-lens-ray-real.gif like this with principal focus at f line on the diagram (2)
1aii)http://mcat-review.org/convex-lens-ray-virtual.gif like this with principal in same place (2)
1bi)use of 1/f-1/u=1/v, which gives -492mm (3)
1bii)magnified, virtual, upright (not inverted) (1)
2a) Quantum efficiency is the percentage of photons hitting the CCD which release electrons (something along those lines) (1)
2bi)0(theta)=l(lambda)/D answer is 1.25x10^-6 rads (i put 1.3 because of S.F.) (1)
2bii)i used s=r0(theta) and got 5AU/10.5LY=7.6x10^-6 rad (3)
2biii) in the exam i misread this, but i believe it is because the planet doesnt emit its own light? (1)
2c)i did: Visible: Transparent to atmosphere, place it anywhere, better in space because atmosphere causes some blur, up to 10m Diameter and blahblahblah
Radio: Transparent to atmosphere, place it anywhere, slightly larger than visible, link them together with seperation of the dishes as the diameter of the lens, blahblah
IR: opaque to atmosphere, absorbed by water vapour, high dry places or space, about same size as visible, needs to be cooled, blahblah (6)
3ai) absolute magnitude is the brightness of a star if it were 10parsecs away. (1)
3aii) i put Betelgeuse(right of table, lower temp) as there is a lower change between the apparent and absolute magnitude. (1)
3bi)lmax(lambda)T=0.0029, gives lmax as 1.3x10^-7m (2)
3bii) black body radiation curve, put scale on x-axis(wavelength) peak is above 1.3x10^-7 (3)
3ci)B (1)
3cii)helium (1)
3ciii)not hot enough, particles dont have enough energy, electrons wont be in n=2 state so hydrogen balmer lines wont be seen. (1)
4a) z=v/c and v=Hd so d=zc/H (convert c into Kms-1 to give 3x10^8) which gives 263. i put 260 (S.F.) and for units Mpc (4)
4b) Radio wave emission/very large redshift (1)

if i got anything wrong, i can change it, just quote me with your opinion.
thanks

I agree with it all except the bold bit, its the fact that the luminosity is so great in my opinion. also with the quantum efficency i sort of agree but prefer, the ratio of detected photons to incident photons, more or less the same.

I've seen this question on other papers, there are several different answers. Radio source would get the marks.

For the Core Section:

Question 2)c)iii) Where you had to work out the value of t, I used my graph from knowing the number of particles was 2*10^22 and read off the graph to work out the time, in years.

My Question is how many marks would this get me, I was lucky because my graph cut the corner of the grid perfectly and I managed to get 2*10^9 years.

When calculated, people have gotten 2.6*10^9 years using N=Noe^-tlambda

I would just like to know if my method is valid, because technically i have "Calculated" t, i just went about it in a practical manner, rather than using a theoretical method.

Hope someone can help me!

For the Core Section:

Question 2)c)iii) Where you had to work out the value of t, I used my graph from knowing the number of particles was 2*10^22 and read off the graph to work out the time, in years.

My Question is how many marks would this get me, I was lucky because my graph cut the corner of the grid perfectly and I managed to get 2*10^9 years.

When calculated, people have gotten 2.6*10^9 years using N=Noe^-tlambda

I would just like to know if my method is valid, because technically i have "Calculated" t, i just went about it in a practical manner, rather than using a theoretical method.

Hope someone can help me!

all i can say is that u are a true engineer! true to the proffession and hopefully a future engineer. but this was a physics exam man. so all i can say is that hope for the best!!!

If they wanted you to use the graph, they would have said "use the graph to estimate a value for t"

Unfortunately you've used the wrong method and that scores no marks.

For the Core Section:

Question 2)c)iii) Where you had to work out the value of t, I used my graph from knowing the number of particles was 2*10^22 and read off the graph to work out the time, in years.

My Question is how many marks would this get me, I was lucky because my graph cut the corner of the grid perfectly and I managed to get 2*10^9 years.

When calculated, people have gotten 2.6*10^9 years using N=Noe^-tlambda

I would just like to know if my method is valid, because technically i have "Calculated" t, i just went about it in a practical manner, rather than using a theoretical method.

Hope someone can help me!

On that basis, you could reject all answers on the basis that they didn't "spoon feed me the method on how to answer the question"

They didn't tell us use a specific formula to calculate t, they just wanted you to calculate it.

If they don't spoon feed you the method, then you need to use the standard way which is to do it theoretically. The question before it was a slight clue, though you could argue against that.

The one thing which is undeniable is that when drawing a curve, it won't ever be perfect. Drawing curves by hand have errors. So in a physics exam you can't use that to find a value as the error is too great.

And the difference between 2 and 2.6 in this case is pretty large.

Who knows what will be on the mark scheme, maybe they may accept it, maybe they won't. Just got to wait and see

I agree with it all except the bold bit, its the fact that the luminosity is so great in my opinion. also with the quantum efficency i sort of agree but prefer, the ratio of detected photons to incident photons, more or less the same.