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OCR Physics B G495 Field and Particle Pictures June 21st 2011 Exam Thread

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100/(2^n)=6.5
Where n = number of half lifes
2^n=(100/6.5)
n*ln2=ln(100/6.5)
n=(ln(100/6.5))/ln2
n= 3.94

Although you are probably right and your method is far more impressive, surely you would just think 1 half life = 50% of value, 2 half lives = 25%, 3 = 12.5% and then 4 = 6.25, which is about 6.5%? It is only 1 mark so i'm hoping i'm not thinking too simple.... :colondollar:

Reply 361

Summerdays

Original post by JoeCarr

You can, but that method is MUCH faster haha :biggrin:

Reply 362

Raimu

How do you find the electric field strength at the midpoint between two charges? I've seen this question a few times and the method was different so I'm not sure how to do it.

Reply 363

Jke

Original post by JoeCarr

Haha, yeah i think that is what you're meant to do, seeing as it says estimate.

Reply 364

wki

http://www.johnbright.conwy.sch.uk/depts/science/Content/Past_Papers/Y13/Physics/2864June2006.pdf
http://www.johnbright.conwy.sch.uk/depts/science/Content/Past_Papers/Y13/Physics/2006JuneMSA2.pdf
I can't get 11 b & ci) ii) ..where does the half life comes up?

(edited 12 years ago)

Reply 365

Summerdays

Original post by Raimu

How do you find the electric field strength at the midpoint between two charges? I've seen this question a few times and the method was different so I'm not sure how to do it.

It's zero if the two charges are equal (and the same polarity.) But it's
E= kq^2/(0.5r)^2 = 4kq^2/r^2 (if the two charges are equal but oppositely charged.)

Reply 366

JoeCarr

Original post by Jke

Haha, yeah i think that is what you're meant to do, seeing as it says estimate.

gotta love the simple methods :biggrin:

Reply 367

Summerdays

Original post by wki

After one half-life there will be 50% lead-206 and 50% Uranium-238, so the ratio is 1:1

But after TWO half-lifes, there will be 25% Uranium-238 and 75% lead-206, so the ratio is now 3:1

After THREE half lifes, there will be 12.5% Uranium-238 and 87.5% lead-206, so the ratio is now 7:1

Hope this helps :smile:

Reply 368

wki

Original post by Birkatron

Januarys raw mark was 74/100 for an A , anyway if i don't post here again until after the exam good luck everyone! :biggrin:

oh yeah cheers mate! good luck! :smile:

Reply 369

Summerdays

Original post by JoeCarr

gotta love the simple methods :biggrin:

Gotta love physics without mathematics :colone:

Reply 370

macleod.mac

Original post by wki

Uranium-238 decays to Lead-206, so whilst the number of U-238 nuclei will fall exponentially, the number of Pb-206 nuclei will increase rapidly then plateau, and as Summerdays has shown, the ratio will therefore change as follows:

After one half life, 50% of U-238 nuclei have decayed to Pb-206, so the ratio is 1:1
After two half lives, 75% of U-238 nuclei have decayed to Pb-206, so the ratio is 1:3

Reply 371

arianex

Original post by Summerdays

Gotta love physics without mathematics :colone:

Summerdays, care to give me a crash course/explanation on the stuff related to our dear friend de Broglie? I've seen a few Qs which ask you to use the dB wavelength to explain/account for a few things (a bit fuzzy now though.)

Cheers :smile:

Reply 372

wki

Original post by Summerdays

oh I didn't notice this. Thanks mate! that really helps :biggrin:

Reply 373

Mikkels88

i was also wondering why sometimes E is about kT, and sometimes its 1.5kT ... what the hell do we use? And how much of internal energy and such should we know? thats physics 4 and thanks to OCR B's infinite genius, i havent revised that yet...

also, anyone done the january 2011 paper? Wondering what people got - i found it really weird and the questions were confusing, managed 86/100 but i went over time to be honest :/

-------------------------------

whoever asked earlier about the atomic and nucleus mass ratio being 3, its because the mass ratio is the same as the volume ratio (same density) and as volume is 4/3 pi r^3 (for a sphere) its by a factor of r^3

Reply 374

Roar.

Is there a list of formulae we need to know that are not in the booklet?

Reply 375

Ollie901

Is the uncertainty stuff even on this paper? I thought that was just G492? There wasnt any in jan anyway.

Reply 376

41jms

guysss what should i do, past papers or rev guide!!

Reply 377

arianex

Original post by 41jms

guysss what should i do, past papers or rev guide!!

Both. Rev guide, then past papers to check what you remember from the rev guide. Lather, rinse, repeat.

Reply 378

Summerdays

Original post by arianex

Sure. His thesis says that the momentum of an object detremines their wavelength, the larger the momentum the smaller the wavelength, but it's really the MASS that determines if this effect is noticable or not.

An electron ha a definite wavelength, depending on its speed. If the wavelength is as large a the diameter of the nucleus, diffreaction occurs, sintheta = 1.2lamda/d where d is the diameter of the nucleus. Each nucleus acts as an obstacle (like an aperture) - interference results between the phasors of the paths available to the electron either side of the nucleus. The theta from this equation determine the angle where a MINIMUM occurs. The electron travel all paths. The smaller the wavelength of an electron, the greater it's resolving abilities (based on rayleigh's criterion, th = lambda/d.) If the energy is large enough an electron can cause the gluon field to stretch so much that mesons are released. An electron is used because it is not effected by the strong force, in the nucleus because it is a lepton :smile:

(edited 12 years ago)

Reply 379

hannahbeex