AQA Physics PHYA5 - Thursday 18th June 2015 [Exam Discussion Thread]

OP

I'm a bit confused;

Can someone explain the different sizes of particles effects neutrons.

So in the control rod neutrons need to be absorbed so does that mean the particles inside the rods are similar sizes or smaller?

Same for the moderator, that needs to withdraw ke from the neutrons, does this mean the moderator has similar sized particles so that elastic collisions can take place and the ke be reduced?

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The control rods simply need to absorb neutrons so, according to Wikipedia...

"Chemical elements with a sufficiently high capture cross-section for neutrons include silver, indium and cadmium. Other elements that can be used include boron, cobalt, hafnium, samarium, europium, gadolinium, terbium, dysprosium, holmium, erbium, thulium, ytterbium, and lutetium.[1] Alloys or compounds of these elements may also be used, such as high-boron steel,[2] silver-indium-cadmium alloy, boron carbide, zirconium diboride, titanium diboride, hafnium diboride, gadolinium nitrate,[3]gadolinium titanate, and dysprosium titanate."

It's so that the element can absorb neutrons without going on to undergo fission itself.

Conversely, the moderator needs to be a similar size molecule so elastic collisions take place. Typically this is graphite (carbon) or water, as these dissipate the kinetic energy of neutrons to thermal levels within about 50 collisions.

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Reply 1801

000alex

2

How much do we need to know about electron scattering to determine nuclear size?

Reply 1802

Aspire101

Original post by gcsestuff

Is this the general idea of the inversion tube experiment:

Lead spheres are heated to a certain temperature.

They are then put in a tube and the tube is rotated a certain number of time.

Due to energy loss from gravity, mgh and the total no. Of rotations we can say that mcT=mghn

Which cancels to give us c= gln/T

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Which bit of the spec is this from? I don't even know it :/

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Reply 1803

CD223

OP

10

Original post by 000alex

How much do we need to know about electron scattering to determine nuclear size?

If you have a look at past six markers (I think it was 2012/2013) then there was one on alpha scattering and electron diffraction. That's the sort of detail :smile:

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Reply 1804

xela238

3

Right, so i had an exam yesterday and beginning to panic. Im going to go though CD223s notes today and then past papers tomorrow, does this sound good?

Reply 1805

Jimmy20002012

4

Original post by CD223

If you have a look at past six markers (I think it was 2012/2013) then there was one on alpha scattering and electron diffraction. That's the sort of detail :smile:

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They have never really done one on the thermal physics bit so should expect that but no idea what they will ask and what you would write since that is not much theory to it!

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Reply 1806

CD223

OP

10

Original post by Jimmy20002012

They have never really done one on the thermal physics bit so should expect that but no idea what they will ask and what you would write since that is not much theory to it!

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Experiment on SHC/SLH maybe?

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Reply 1807

Jimmy20002012

4

Original post by CD223

Experiment on SHC/SLH maybe?

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What would we even write for that?

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Reply 1808

CD223

OP

10

Original post by Jimmy20002012

What would we even write for that?

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Apparatus, measurements, plotting of results on a graph, conclusions from graph/calculations etc. there's plenty to say.

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Reply 1809

ZedRigby

Hi everyone,

I don't study maths and therefore am not entirely sure on how to manipulate logs in a formula. Would someone who is studying astrophysics be able to post all of the variations of the formula m-M = 5log d/10 for me? It would help me out a lot :smile:

Reply 1810

CD223

OP

10

Original post by ZedRigby

Hi everyone,

I don't study maths and therefore am not entirely sure on how to manipulate logs in a formula. Would someone who is studying astrophysics be able to post all of the variations of the formula m-M = 5log d/10 for me? It would help me out a lot :smile:

Does this help?

[br]m = 5 \log \left( \dfrac{d}{10} \right) + M[br]

[br]M = m - 5 \log \left( \dfrac{d}{10} \right)[br]

[br]d = 10 \times 10^{ \left( \frac{m-M}{5} \right)}[br]

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Reply 1811

toddle1

1

Original post by ZedRigby

Hi everyone,

I don't study maths and therefore am not entirely sure on how to manipulate logs in a formula. Would someone who is studying astrophysics be able to post all of the variations of the formula m-M = 5log d/10 for me? It would help me out a lot :smile: