AQA A Level Physics Paper 1 UNOFFICIAL MARK SCHEME (20th May 2019)

Updated link with corrections.

Reply 382

mayaaaaaahhhhh

8

You’re a literal god

Original post by bozdag

You're all welcome.

https://drive.google.com/open?id=17AmgKhqTIXfjSsktzaj7gEIZhY6eJLMI

Reply 383

Ckweekes

2

I think I’ve lost 20/25 marks in this paper... ordinarily that’s good but considering everyone seems to have done well do you think the grade boundaries will be really high???

Reply 384

Hello1212122434

6

How did people in you're school or college find it?
What mark do you think you got?

Original post by oly1357

You reckon these will be the grade boundaries? They seem quite high relative to the last few years exam papers

Reply 385

Original post by Ckweekes

I think I’ve lost 20/25 marks in this paper... ordinarily that’s good but considering everyone seems to have done well do you think the grade boundaries will be really high???

Last year you could drop 23 marks n still get an A* so grade boundaries will most likely be similar to that. Twitter was freaking out about the paper so I think most of the country struggled so you'll be fine.

Reply 386

People thought it was an odd paper since massive topics like quantum phenomena and standing waves were left out. Most people found it fine tho. I think I got about 78/85 minus some for poor communication.

Original post by Hello1212122434

How did people in you're school or college find it?
What mark do you think you got?

Reply 387

Agzamkrut

Thank you mate! Is “ray perpendicular to the surface of the prism” would be the right thing to write? Because that is what I wrote, because everyone wrote that it is normal to the line.

Original post by bozdag

You're all welcome.

https://drive.google.com/open?id=17AmgKhqTIXfjSsktzaj7gEIZhY6eJLMI

Reply 388

bozdag

Means the same thing.

Original post by Agzamkrut

Thank you mate! Is “ray perpendicular to the surface of the prism” would be the right thing to write? Because that is what I wrote, because everyone wrote that it is normal to the line.

Reply 389

zaid39401

Thicker diameter in one of the wires = greater cross sectional area, resistivity = RA/L so surely the resistivity will be greater for the thicker wire? It can't be the same for wires of different diameters

Original post by bruffsupporter

Quantities the same in both wires - resistivity and current (B)

Reply 390

username2912782

13

Original post by zaid39401

Thicker diameter in one of the wires = greater cross sectional area, resistivity = RA/L so surely the resistivity will be greater for the thicker wire? It can't be the same for wires of different diameters

Resistivity is the same for a fixed material as it's a bulk property. (And they're made of the same material)

(edited 4 years ago)

Reply 391

Resistance of the two cylinders is also different tho so that's why resistivity is constant.

Original post by zaid39401

Thicker diameter in one of the wires = greater cross sectional area, resistivity = RA/L so surely the resistivity will be greater for the thicker wire? It can't be the same for wires of different diameters

Reply 392

unknown_hipster

9

Resistivity is the material's property, resistance is the reaction (that's probably not the word to describe it but I couldn't think of anything else)

Original post by CianBlake

Resistance of the two cylinders is also different tho so that's why resistivity is constant.

Reply 393

Duck3E

Could you have talked about beta minus being a weak interaction and it having the exchange particle of a W boson, while for the internal one whatever it was called is electromagnetic so you have a photon as the exchange particle

Reply 394

douglas.morgan

6

Another few points for the internal conversion question may be:

1. The constituent nucleons of the nucleus (i.e. the proton and neutron numbers) do not change in internal conversion.
2. The emitted electron in internal conversion is not referred to as a beta(-minus) particle as it is not the product of beta-decay.
3. For internal conversion, the energy of the emitted electron follows a discrete energy spectrum (energy of electrons can from internal conversion can only be certain values) but beta-particles follow a continuous energy distribution.

or the converse of each statement

Reply 395

Manlikeytu

12

Original post by Duck3E

Could you have talked about beta minus being a weak interaction and it having the exchange particle of a W boson, while for the internal one whatever it was called is electromagnetic so you have a photon as the exchange particle

that's what I said

Reply 396

DontSweat

8

Just wondering as I did get the first two, but for 3 I said that eventually, internal conversion will result in an orbiting electron falling back down an energy level to replace that which was liberated, releasing a photon of energy equal to that of the energy level change. Whereas Beta Minus would not result in photon emission.

Original post by douglas.morgan

Another few points for the internal conversion question may be:

1. The constituent nucleons of the nucleus (i.e. the proton and neutron numbers) do not change in internal conversion.
2. The emitted electron in internal conversion is not referred to as a beta(-minus) particle as it is not the product of beta-decay.
3. For internal conversion, the energy of the emitted electron follows a discrete energy spectrum (energy of electrons can from internal conversion can only be certain values) but beta-particles follow a continuous energy distribution.

or the converse of each statement

Reply 397

Original post by Duck3E

Could you have talked about beta minus being a weak interaction and it having the exchange particle of a W boson, while for the internal one whatever it was called is electromagnetic so you have a photon as the exchange particle

This is definitely a mark

Reply 398