PHYA5 ~ 20th June 2013 ~ A2 Physics

Reply 1640

bugsuper

8

Original post by SpiggyTopes

Yes you're right, sorry. I must have misread what you said, I thought you said 24.7 seemed too high.

that's all right, I thought that like 200MeV or whatever someone else said seemed too high. sorry for the confusion

Reply 1641

Acruzen

10

Original post by bugsuper

FISSION releases more energy than FUSION
because even though the change in binding energy for each nucleon is less there are more nucleons involved

oh yeah. i was thinking about energy per kg of fuel

Reply 1642

Kenna

I personally thought that the June 2012 with turning points was easier than this paper.

Reply 1643

sports_crazy

Original post by bugsuper

FISSION releases more energy than FUSION
because even though the change in binding energy for each nucleon is less there are more nucleons involved

I am confused. Is my answer 24.7 correct or not. Did they ask energy per nucleon or per nucleus

Reply 1644

mar junior

This paper was very bad for me.. sort of gave up because i couldnt do any of question 3..

For astrophysics did anyone else get 71.6 OR 72 For hubbles constant?

Reply 1645

FloydHead

10

Original post by posthumus

de Broglie one .... 4 marker. I thought it was pretty sneaky

did anyone convert the eV into V by multiply eV by e

then multiplying it by e again to get eV

I thought it was sneaky as hell

Wait, what? I converted eV to joules, then used the kinetic energy approximation formula (I'm still annoyed they didn't tell us to ignore relativistic effects, they should be clearer) to find momentum and then put that into the de Broglie equation. You can't convert electronvolts to volts, electronvolts is a unit of energy, volts isn't.

Reply 1646

cooldudeman

18

Original post by Kenna

I personally thought that the June 2012 with turning points was easier than this paper.

Really? The section A in jun12 was a bit harder but the turning points were the same.

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Reply 1647

fizzbizz

4

Guys from what I remember, the equation was:

4H -> He + 4e + ve

Mass of the helium nucleus was 4.002602u
The hydrogen atom was just a proton

So mass before = 4 x 1.67 x 10^-27 kg
Mass after = (4.002602 x 1.661 x 10^-27) + (4 x 9.11 x 10^-31) kg

Difference in mass = 2.8 x 10^-29 kg = 0.016878u

0.016878 x 931.3 = 15.7 MeV ?!!!

Can someone explain what I'm doing wrong

Reply 1648

Dslash

Original post by FloydHead

Wait, what? I converted eV to joules, then used the kinetic energy approximation formula (I'm still annoyed they didn't tell us to ignore relativistic effects, they should be clearer) to find momentum and then put that into the de Broglie equation. You can't convert electronvolts to volts, electronvolts is a unit of energy, volts isn't.

I did this, i'm pretty sure my db-wavelength was 2.0x10^-(something..) to 2 sig figs.

Reply 1649

fayled

9

Defined the atomic mass unit as 1/12th of the mass of a carbon-12 nucleus. Think I should have wrote atom instead...

Reply 1650

Infinity1

Original post by fizzbizz

Guys from what I remember, the equation was:

4H -> He + 4e + ve

Mass of the helium nucleus was 4.002602u
The hydrogen atom was just a proton

So mass before = 4 x 1.67 x 10^-27 kg
Mass after = (4.002602 x 1.661 x 10^-27) + (4 x 9.11 x 10^-31) kg

Difference in mass = 2.8 x 10^-29 kg = 0.016878u

0.016878 x 931.3 = 15.7 MeV ?!!!

Can someone explain what I'm doing wrong

Wasn't it two positrons, not four?

Reply 1651

josephtsui

Original post by fizzbizz

Guys from what I remember, the equation was:

4H -> He + 4e + ve

Mass of the helium nucleus was 4.002602u
The hydrogen atom was just a proton

So mass before = 4 x 1.67 x 10^-27 kg
Mass after = (4.002602 x 1.661 x 10^-27) + (4 x 9.11 x 10^-31) kg

Difference in mass = 2.8 x 10^-29 kg = 0.016878u

0.016878 x 931.3 = 15.7 MeV ?!!!

Can someone explain what I'm doing wrong

1) there are 2 positrons, not 4
2) the mass is not 9.11 x 10^-31, that's in kg, you want it to be in u (atomic mass unit), which is 5.5 x 10^-4 u (i think)

Reply 1652

OmegaKaos

Original post by fizzbizz

Guys from what I remember, the equation was:

4H -> He + 4e + ve

Mass of the helium nucleus was 4.002602u
The hydrogen atom was just a proton

So mass before = 4 x 1.67 x 10^-27 kg
Mass after = (4.002602 x 1.661 x 10^-27) + (4 x 9.11 x 10^-31) kg

Difference in mass = 2.8 x 10^-29 kg = 0.016878u

0.016878 x 931.3 = 15.7 MeV ?!!!

Can someone explain what I'm doing wrong

Why are you converting to kg then back to atomic units.
just use atomic all the way through and you would of got the correct value of 24.7mev

Reply 1653

fayled

9

Original post by AWithers95

Having done "Turning Points", I can't comment on the astro q's...
I agree with your answers - I got 0.016kg for the ice q.

Same for the ice...

Reply 1654

fizzbizz

4

Original post by Infinity1

Wasn't it two positrons, not four?

Original post by josephtsui

1) there are 2 positrons, not 4
2) the mass if not 9.11 x 10^-31, that's in kg, you want it to be in u (atomic mass unit), which is 5.5 x 10^-4 u (i think)

Even with two positrons, I get 16.7 MeV?!

And I'm aware of that, which is why I converted everything to kg first, then converted the mass difference to atomic mass units

Reply 1655

JamesSS95

Original post by fizzbizz

Guys from what I remember, the equation was:

4H -> He + 4e + ve

Mass of the helium nucleus was 4.002602u
The hydrogen atom was just a proton

So mass before = 4 x 1.67 x 10^-27 kg
Mass after = (4.002602 x 1.661 x 10^-27) + (4 x 9.11 x 10^-31) kg

Difference in mass = 2.8 x 10^-29 kg = 0.016878u

0.016878 x 931.3 = 15.7 MeV ?!!!

Can someone explain what I'm doing wrong

Not sure if I'm right but I used the numbers for rest mass energy on the second page of the data booklet.

(4(938.257))-((4.002602)(931.3)+2(0.510999))=24.38MeV

I checked on the internet for the equation and a website said it was ~27MeV so I think that's the right sort of answer

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Reply 1656

bugsuper

8

Original post by fizzbizz

Guys from what I remember, the equation was:

4H -> He + 4e + ve

Mass of the helium nucleus was 4.002602u
The hydrogen atom was just a proton

So mass before = 4 x 1.67 x 10^-27 kg
Mass after = (4.002602 x 1.661 x 10^-27) + (4 x 9.11 x 10^-31) kg

Difference in mass = 2.8 x 10^-29 kg = 0.016878u

0.016878 x 931.3 = 15.7 MeV ?!!!

Can someone explain what I'm doing wrong

the mass after wasn't that. It was 4.00150, wasn't it? because it was a helium-4 nucleus, which is an alpha particle

Reply 1657

bugsuper

8

i guess it depends on what the mass of the final nucleus was, but I thought it was 4.00150u

Reply 1658

fizzbizz

4

Original post by bugsuper

the mass after wasn't that. It was 4.00150, wasn't it? because it was a helium-4 nucleus, which is an alpha particle

http://uk.ask.com/question/helium-nucleus

Reply 1659

masryboy94

8

that was a nice paper ! and the medical physics section was even nicer !!!!! ahhhhh !! THANK THE LORD !!! revision really has paid off, hope you all smashed your examsss and get into the uni's you want :biggrin:

THANK YOU ALL FOR YOUR HELPPP, SERIOUSLY COULDN'T HAVE DONE IT WITHOUT YOU LOT :biggrin:

PHYA5 ~ 20th June 2013 ~ A2 Physics

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