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AQA Physics PHYA5 - 28th June 2016 [Exam Discussion Thread]

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Original post by marioman
The count rate is going to drop when you move it further away regardless of whether there's beta emission or not.


Wouldnt there be a rapid drop past beta's range?
Original post by Nikhilm
Nah it was 127/128 from the calculation


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Yeah!
Original post by CourtlyCanter
I was the only one in my school to do it as I learnt it myself because the teacher we had for astrophysics was awful. This year's paper was hard. Do you remember your answers to any of the questions?


I've made a medical thread
Original post by Mango Milkshake
just checked, it's 10,000



did you round it too early?

I didn't round anything off until I got to the final answer. Don't know where I went wrong. :frown:
Original post by txnilxnur
I thought it was 2000 lmao, swear that's what it says in the textbook


Isn't it 1 in 2000 for a deflection and 1 in 10000 for angles greater than 90?

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Original post by CourtlyCanter
I was the only one in my school to do it as I learnt it myself because the teacher we had for astrophysics was awful. This year's paper was hard. Do you remember your answers to any of the questions?


yea this paper was pretty difficult. I didnt get the first calculation to do with the eye. I remember getting 57% for the x ray calculation to do with 12mm. what about you?
Original post by lc970916
probably not .

" Expressions for the electric and magnetic fields in free space contain the electric permittivity ε0 and magnetic permeability μ0 of free space. As indicated in the section on electric and magnetic constants, these two quantities are not independent but are related to "c", the speed of light and other electromagnetic waves."

source: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elefie.html


True they are well defined quantities rather than those who put electric foeld strength and magnetic field strength which are not defined quantities.
3.1*10^-15 for radius of oxygen


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Original post by particlestudent
@kingaaran @Mango Milkshake

What did you guys put for the test to see if it emitted beta as well?


Use a beta absorber, i.e. a few mm of aluminium. Then you can see if this result matches up with the result obtained at 0.9m by using the inverse square law of gamma radiation
Original post by Nikhilm
Ya so you have to use aluminium sheet to check, or some other thing


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Could I have said:

"I said plot a graph of count rate against 1/r^2 and it should be a straight line if it is only gamma, but if it is beta it will not be a straight line."
specific charge was 1.8x10^11 Ckg^-1
Original post by BigDaddy97
3.1*10^-15 for radius of oxygen


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got that
Original post by FireBLue97
specific charge was 1.8x10^11 Ckg^-1


It wasn't. That is the actual specific charge but the experimental specific charge was 9.1x10^10
Original post by kingaaran
Use a beta absorber, i.e. a few mm of aluminium. Then you can see if this result matches up with the result obtained at 0.9m by using the inverse square law of gamma radiation


I didn't mention aluminium at all :frown:. I said plot count rate against 1/distance squared and if it is a complete straight line then it is only a gamma emitter
Grade boundary predictions?


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Reply 775
Original post by particlestudent
@kingaaran @Mango Milkshake

What did you guys put for the test to see if it emitted beta as well?


I put put an electromagnet between source and detector and drew this ImageUploadedByStudent Room1467110577.548067.jpg


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Original post by particlestudent
I didn't mention aluminium at all :frown:. I said plot count rate against 1/distance squared and if it is a complete straight line then it is only a gamma emitter


I think you'll be fine - it sounds valid
Original post by particlestudent
I said plot a graph of count rate against 1/r^2 and it should be a straight line if it is only gamma, but if it is beta it will not be a straight line.

This better be accepted :angry:

I saw that in a past paper as an alternative method once. You still get the marks
I also got 1.2x10^8 AU for the penultimate quasar question but starting to think this is wrong after looking on here :frown:

For those who got the other answer (2.4 something) could you tell me how u did this question as i am really confused.
Original post by cowie
I put put an electromagnet between source and detector and drew this ImageUploadedByStudent Room1467110577.548067.jpg


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That is clever :wink: Definitely going to get credit there

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