AQA Physics A Unit 5 Revision Thread - 27th June

Ok could someone shed some light on this for me.
On the stability curve are the alpha emitters below the stable nuclei? Because that what's I was taught because they have too many protons/ are too heavy so realso some in the form of alpha.
But in the nelson thornes text book it says that alpha emitters occur above the stable nuclei. So now I'm confused.
Does anyone know which one is right?

Nuclei emit alpha particles because they are too heavy and not because of the proton/neutron ratio, so alpha emitters are both above and below the line of stability.
Alpha emitters are only ever heavy nuclei though, so alpha emitters only occur after z = 82 (anything with 82 or more protons is potentially an alpha emitter).

Beta + occurs below the line an B- occurs above the line. The line is linear until N=20 and Z=20, at which point it curves upwards. The only other thing you might get asked is to draw an arrow representing a decay, but I think that's all you need to know about that graph.

Draw a diagram to help. If Io is in orbit around Jupiter at a radius of 4.5 x 10^5 km and the distance from the Earth to Jupiter is 6.0 x 10^8 km then you get a triangle. The angluar separation, theta is the angle subtended by the line from the Earth to Io and from the Earth to Jupiter.

So tan theta= 4.5 x 10^5 / 6.0 x 10^8. Using the small angle approximation you don't need to include the tan bit, just divide the numbers. You get 7.5 x 10^-4. This is the angular separtion when viewed with the unaided eye. The telescope magnifies the angle by a factor of 30, so just multiple by 30 and the answer is 0.021 rad.

If there's a question on controlling the rate of fission in a reactor, is it best to reference the moderator or the control rods?

And what exactly is the difference between them in terms of purpose?

If there's a question on controlling the rate of fission in a reactor, is it best to reference the moderator or the control rods?

And what exactly is the difference between them in terms of purpose?

Its at the centre of curvature of the mirror

Isn't that assuming that all the air in the container goes into the pump?

Could someone give me a full proof method of working out how to work out a new pressure of a container if a pump sucks it out ?

Alpha emitters are above z=60.

I just spent about 15 mins looking at the question you mentioned from the specimin and could not for the life of me work out how they got the answer in the mark scheme (83.5 kPa). The working they showed was

p = 99 × 3.50/4.15

but I can't work out where the 4.15 came from, ususally you can work backwards and at least see what calculation they did. Could this be another specimin mistake? This was my working:

p1 x V1 =p2 x V2
99 x 3.5 x10^-4 = p2 x (3.5 + 6.5)x10^-4
p2 = 34.7 kPa

I don't know if you've got this result when doing it as well but these are usually pretty simple and if we're not getting it then there could be something wrong with the paper or mark scheme.

But I would probably word it better than that :P

I'm sat with a periodic table in front of me from Chemistry. Tungsten, platinum, gold, mercury, lead and neodymium all have a proton number above 60,and none are alpha emitters.
It is 82. Bismuth (83) is radioactive if I remember correctly, and Polonium (84) certainly is, as it was one of the first radioactive elements discovered.
Alpha emmiters are to the right of Z=82 (lead has a proton no. of 82 and I don't think lead is radioactive).

Ok, well my text book says above z=60. Also, consider the curve they ask you to draw for the line of stability. They only want it up to z=80, so you wouldn't be able to mark on any alpha emitters.

"The smallest nuclei which have to date been found to be capable of alpha emission are the lightest nuclides of tellurium (element 52), with mass numbers between 106 and 110" - http://alpha-particle.co.tv/

I just spent about 15 mins looking at the question you mentioned from the specimin and could not for the life of me work out how they got the answer in the mark scheme (83.5 kPa). The working they showed was

p = 99 × 3.50/4.15

but I can't work out where the 4.15 came from, ususally you can work backwards and at least see what calculation they did. Could this be another specimin mistake? This was my working:

p1 x V1 =p2 x V2
99 x 3.5 x10^-4 = p2 x (3.5 + 6.5)x10^-4
p2 = 34.7 kPa

I don't know if you've got this result when doing it as well but these are usually pretty simple and if we're not getting it then there could be something wrong with the paper or mark scheme.

Ok, well my text book says above z=60. Also, consider the curve they ask you to draw for the line of stability. They only want it up to z=80, so you wouldn't be able to mark on any alpha emitters.

"The smallest nuclei which have to date been found to be capable of alpha emission are the lightest nuclides of tellurium (element 52), with mass numbers between 106 and 110" - http://alpha-particle.co.tv/

He speaks the truth.

Go to post a few up, apparently you miscopied the numbers Right method though

Yeah, that's my method too... As the volume of the system is effectively increasing, so the new pressure must be less I think the mark scheme is wrong, as that's the right method I think

You're problem is that the 6.5x10^-4 you typed, is actually x10^-5
So it's 3.5x10^-4 + 6.5x10^-5
giving 4.15x10-4