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

Original post by betbi3etwerrd

1. Calculate the change in binding energy (short) not account for neutrons since they have no binding energy

Not sure about this

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Reply 1662

Original post by CD223

It's the way log rules work - don't worry if it confuses you, but you can cancel the minus sign by flipping the fraction inside the "ln" brackets.

If you want I can provide a more detailed explanation as to why this works, but by the looks of things you get the correct answer your way anyway and won't lose marks for dividing by the negative number instead.

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Oh ok so the minus makes it work... That's cool that's all I need to know.... I would love to understand why but I doubt I would understand an explanation of it lol and at this moment in time I've got other things to learn. Thanks though 😊

Reply 1663

Original post by a.a.k

Not sure about this

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I've got multiple sources approving this and the mark scheme does it this way for that question.

read pg188 Nelson Thornes AQA A2 physics
paragraph starting with "Energy is released.."

or just see that the mark scheme worked out the change in binding energy : http://filestore.aqa.org.uk/subjects/AQA-PHYA5-1-W-MS-JUN14.PDF

By the way I said the binding energy of neutrons are zero but this is obviously only true for the neutrons that are not in a nucleus.

(edited 8 years ago)

Reply 1664

Original post by CD223

So essentially it doesn't matter which method you use. Simply use the quicker one based one what you're given in the question?

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Well it didn't work for most people who used the mass defect because they were calculating it wrong.

For this question to find the mass of a nucleus , you had to add up the mass of what it'd be if the neutrons and protons were apart then subtract this value from the binding energy of the nucleus (since by definition binding energy of nucleus is the energy needed to pull a part a nucleus into it's constituent protons and neutrons).

Do that for each nucleus and then find the change in mass, AKA mass defect, (also account for the neutron masses).

OR you could simply find the change in binding energy (where the binding energy of neutrons are 0 since they're already constituent.

(edited 8 years ago)

Reply 1665

ubisoft

Are you ever asked to calculate the rms speed, or is it always given?

Reply 1666

Original post by ubisoft

Are you ever asked to calculate the rms speed, or is it always given?

I'm sure they could easily get us to do that

Reply 1667

http://filestore.aqa.org.uk/subjects/AQA-PHYA5-1-QP-JUN13.PDF

HELPPP!!! Q.1 (c) ii)

heres the mark scheme: http://filestore.aqa.org.uk/subjects/AQA-PHYA5A-W-MS-JUN13.PDF

Why subtract the two electron mass not add?

Reply 1668

CD223

OP

10

Original post by GraceYux

http://filestore.aqa.org.uk/subjects/AQA-PHYA5-1-QP-JUN13.PDF

HELPPP!!! Q.1 (c) ii)

heres the mark scheme: http://filestore.aqa.org.uk/subjects/AQA-PHYA5A-W-MS-JUN13.PDF

Why subtract the two electron mass not add?

You want to find the difference in mass between reactants and products. By taking the total mass of the products away from the reactant you get the mass defect in atomic mass units, which you can multiply by 931.3MeV to get the energy released from the process.

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Reply 1669

Sumz.96

1

This question is from June 2011 ASTRO

2 (b) (ii) The power output of Deneb is 70000 times greater than the Sun. Calculate the radius of Deneb.
surface temperature of the Sun = 5700 K
black body temperature of Deneb is 8.5x10^3K

Please help. I don't understand the last step in the mark scheme.

Reply 1670

Original post by CD223

You want to find the difference in mass between reactants and products. By taking the total mass of the products away from the reactant you get the mass defect in atomic mass units, which you can multiply by 931.3MeV to get the energy released from the process.

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But surely to find the total mass of the products, you would add the mass of the He nucleus with the two electrons released?

Reply 1671

CD223

OP

10

Original post by GraceYux

But surely to find the total mass of the products, you would add the mass of the He nucleus with the two electrons released?

Yes, then subtract that figure from the mass of the 4 hydrogen nuclei (what I termed as the "reactants").

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Reply 1672

Original post by GraceYux

http://filestore.aqa.org.uk/subjects/AQA-PHYA5-1-QP-JUN13.PDF

HELPPP!!! Q.1 (c) ii)

heres the mark scheme: http://filestore.aqa.org.uk/subjects/AQA-PHYA5A-W-MS-JUN13.PDF

Why subtract the two electron mass not add?

This is simply a mass defect question. From the binding energy per nucleon curve you can probably estimate that the binding energy per nucleon of Hydrogen (1 proton 0 neutron) is 0 MeV.
Which basically means that you can take the mass of a proton for the Hydrogen.

Do everything in atomic mass units (u) as in the formula book it gives you the masses of electrons, neutrons and protons in atomic mass units.

This makes it easier and at the end you simply multiply your answer by 931.5 to get MeV.

SO to asnwer your question:

find TOTAL mass of left hand side (hydrogen)
find TOTAL mass of right hand side (helium and electron)
find the difference in these masses (hydrogen - (helium + electron)) = (hydrogen - helium - electron)
give answer in MeV

(edited 8 years ago)

Reply 1673

Original post by betbi3etwerrd

This is simply a mass defect question. From the binding energy per nucleon curve you can probably estimate that the binding energy per nucleon of Hydrogen (1 proton 0 neutron) is 0 MeV.
Which basically means that you can take the mass of a proton for the Hydrogen.

Do everything in atomic mass units (u) as in the formula book it gives you the masses of electrons, neutrons and protons in atomic mass units.

This makes it easier and at the end you simply multiply your answer by 931.5 to get MeV.

SO to asnwer your question:

find TOTAL mass of left hand side (hydrogen)
find TOTAL mass of right hand side (helium and electron)
find the difference in these masses (hydrogen - (helium + electron)) = (hydrogen - helium - electron)
give answer in MeV

but its not '(hydrogen - (helium + electron))' , its 'H - (He - electron)

Reply 1674

Original post by GraceYux

but its not '(hydrogen - (helium + electron))' , its 'H - (He - electron)

no it's not

it is hydrogen - helium - electron

which is the same as hydrogen - (helium + electron)

Reply 1675

Original post by betbi3etwerrd

no it's not

it is hydrogen - helium - electron

which is the same as hydrogen - (helium + electron)

Oops, I'm an idiot, I get it now, thank you!!

Reply 1676

aprocrastinator

10

apparently a van got raided into so loads of past papers had to be rewritten by AQA including Physics Unit 5 :O so I'm expecting some weird questions, probably with mistakes in...

Reply 1677

Original post by aprocrastinator

apparently a van got raided into so loads of past papers had to be rewritten by AQA including Physics Unit 5 :O so I'm expecting some weird questions, probably with mistakes in...

I don't get the point of why anyone would do this... It's not like people wouldn't notice all the papers missing.... Anyway you'd think by now aqa would have learnt from the past and create a backup paper for if this happens

Reply 1678

RobHunter97

Original post by jf1994

Nah it says real object not real image

Wow I'm stupid. Need to read the questions properly....

Thank you

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Reply 1679