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OCR Physics A Paper 3 Unified Physics 3rd June 2019 Predictions

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What did people get for the electron shell arrow, i put my arrow up, cant remember the value
Reply 61
Avatar for Jeff117
Jeff117
1
Arrow had to be down as a photon was being emitted, not absorbed
Original post by ElbertAinstein
What did people get for the electron shell arrow, i put my arrow up, cant remember the value
Hopefully it goes up soon, honestly so worried on how this exam went
Original post by Montytheragdoll
Any unofficial mark scheme threads?
same but i think the grade boundaries will be fairly low for paper 3. IMO i found it harder than last years paper 3
Original post by RIghty Ross
Hopefully it goes up soon, honestly so worried on how this exam went
****ing solid paper
Definitely, I think that was harder than 2017. I completely messed up this paper.
Original post by Zaki1039
same but i think the grade boundaries will be fairly low for paper 3. IMO i found it harder than last years paper 3
Yeah me too. I just want to see what everyone else got
Original post by Zaki1039
same but i think the grade boundaries will be fairly low for paper 3. IMO i found it harder than last years paper 3
I thought since a photon was being emitted, the electron would go from a higher negative energy to a lower negative energy
Original post by Jeff117
Arrow had to be down as a photon was being emitted, not absorbed
Nah because the electron loses energy as it emits the photon
Original post by ElbertAinstein
I thought since a photon was being emitted, the electron would go from a higher negative energy to a lower negative energy
Down arrow between the levels with 4.09x10^-19 difference if I remember correctly
Original post by Montytheragdoll
Nah because the electron loses energy as it emits the photon
I think I've ruined my grade with paper 3. Could have gotten an A but I didn't answer like half this paper
Yeah that's the 6 marker I'm happy about, it was just a really straightforward plonk the numbers into lambda = ax/D
Original post by Josh mcbeth
I never want to see a torch for the rest of my life... Everything else in the paper I thought was straight forward. The diffraction grating 6 marker was horribly worded but the numbers were easy to follow Apart from that though it was ok 🙂
F**king **** torch
I found it quite challenging, may messed up a few Qs but nothing I found impossible.

I got 667nm I think, for the 2nd 6 marker. Also I got 30 something for the initial acceleration (may have got wrong though as I had 20 sec to do) and 50 something for the constant from graph.
(edited 4 years ago)
Original post by roryjamesallen
Yeah that's the 6 marker I'm happy about, it was just a really straightforward plonk the numbers into lambda = ax/D


Hmmm surely it’s dsin theta = n lambda because it was a diffraction grating
You don't use that. This is a diffraction grating so you use nlamda=dsintheta. With the values from ruler, you do trig to calucalte sin theta
Original post by roryjamesallen
Yeah that's the 6 marker I'm happy about, it was just a really straightforward plonk the numbers into lambda = ax/D
I got both the same answers
Original post by RedGiant
I found it quite challenging, may messed up a few Qs but nothing I found impossible.

I got 667nm I think, for the 2nd 6 marker. Also I got 30 something for the initial acceleration (may have got wrong though as I had 20 sec to do) and 50 something for the constant from graph.
Reply 77
Avatar for Dh3van
Dh3van
7
did you have to work it out? it didnt say to but i didnt know the percentage unceratinty but the explanation was easy pythagoras and stuff


Original post by RedGiant
I found it quite challenging, may messed up a few Qs but nothing I found impossible.

I got 667nm I think, for the 2nd 6 marker. Also I got 30 something for the initial acceleration (may have got wrong though as I had 20 sec to do) and 50 something for the constant from graph.
Original post by roryjamesallen
Yeah that's the 6 marker I'm happy about, it was just a really straightforward plonk the numbers into lambda = ax/D


you're meant to use dsin(theta) = n landa
the vertical acceleration is 298ms-2

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