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    (Original post by -jordan-)
    I think it was 3 or 4 marks, I think you'll only get one. You should get e.c.f for the number of mols if that's the method you used
    Im almost sure it was 2 marks like haha
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    (Original post by allofthestars)
    Im almost sure it was 2 marks like haha
    Can confirm it was 2 was the specific charge 1.8x10^11 Ckg^-1
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    (Original post by FireBLue97)
    Can confirm it was 2 was the specific charge 1.8x10^11 Ckg^-1
    I got 1.9x10^11 Ckg^-1

    Did you use \dfrac{e}{m} = \dfrac{2V}{B^2r^2}?
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    For the question asking about the population experiencing a higher background radiation and the possible causes for it - would the answer of "Radioactive building materials used in construction" got the mark?

    It is an answer in the text book however I don't know if it's relevant to the "in the past 100 years" part
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    (Original post by js.int)
    I got 1.9x10^11 Ckg^-1

    Did you use \dfrac{e}{m} = \dfrac{2V}{B^2r^2}?
    Do you remember what the numbers were for V, r and B? Isn't the two supposed to be on the bottom for that formula?
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    (Original post by -jordan-)
    Do you remember what the numbers were for V, r and B?
    Nope but I remember it the specific charge to be something like 1.87 or 1.89 x 10^11 Ckg^-1
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    (Original post by -jordan-)
    ... Isn't the two supposed to be on the bottom for that formula?
    Nope:

    

E_k = eV = \dfrac{1}{2}mv^2, v = \dfrac{Ber}{m}



\therefore eV = \dfrac{1}{2}m(\dfrac{Ber}{m})^2



\therefore eV = \dfrac{1}{2}(\dfrac{B^2e^2r^2}{m  })



\therefore V = \dfrac{1}{2}(\dfrac{B^2er^2}{m})



\therefore 2V = \dfrac{B^2er^2}{m}



\therefore \dfrac{e}{m} = \dfrac{2V}{B^2r^2}
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    (Original post by js.int)
    Nope:

    

E_k = eV = \dfrac{1}{2}mv^2, v = \dfrac{Ber}{m}



\therefore eV = \dfrac{1}{2}m(\dfrac{Ber}{m})^2



\therefore eV = \dfrac{1}{2}(\dfrac{B^2e^2r^2}{m  })



\therefore V = \dfrac{1}{2}(\dfrac{B^2er^2}{m})



\therefore 2V = \dfrac{B^2er^2}{m}



\therefore \dfrac{e}{m} = \dfrac{2V}{B^2r^2}
    Hm that's odd. Antonine has the two on the bottom and I got a two by using Bev=eV/d and replacing d by 2r...
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    (Original post by -jordan-)
    Hm that's odd. Antonine has the two on the bottom and I got a two by using Bev=eV/d and replacing d by 2r...
    Isn't \dfrac{eV}{d} for uniform electric fields?... not magnetic fields?
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    (Original post by js.int)
    Isn't \dfrac{eV}{d} for uniform electric fields?... not magnetic fields?
    The question had both an electric field and a magnetic field, you were told the potential at which it's accelerated.

    1/2mv^2 = eV is for a uniform electric field, using Work Done = QV. I used the force instead rather than work done. There is some reason why my method doesn't work I just can't think why.
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    (Original post by -jordan-)
    The question had both an electric field and a magnetic field, you were told the potential at which it's accelerated.

    1/2mv^2 = eV is for a uniform electric field, using Work Done = QV. I used the force instead rather than work done
    But surely once an electron has been accelerated and has left the anode, it isn't in the electric field any more? So there is no electric force on the electron...

    Say, if in fact, there is still an electric force on the electrons, wouldn't it cause the electron to accelerate? (Tangental to the circle it follows) ... and hence cause the radius of the circular beam to increase?
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    (Original post by js.int)
    But surely once an electron has been accelerated and has left the anode, it isn't in the electric field any more? So there is no electric force on the electron...

    Say, if in fact, there is still an electric force on the electrons, wouldn't it cause the electron to accelerate? (Tangental to the circle it follows) ... and hence cause the radius of the circular beam to increase?
    Yes, you are correct. I still got V/B^2r^2 anyway, so I have half the expected value, so hopefully only dropped a mark. Was it two?
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    (Original post by -jordan-)
    Yes, you are correct. I still got V/B^2r^2 anyway, so I have half the expected value, so hopefully only dropped a mark. Was it two?
    I think it was 4 marks for that question because it was meant to be quite a long derivation
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    (Original post by js.int)
    I think it was 4 marks for that question because it was meant to be quite a long derivation
    Marks were surely not for derivation though, it's given in the books and one you can just learn.
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    (Original post by -jordan-)
    Marks were surely not for derivation though, it's given in the books and one you can just learn.
    I don't know tbh... I remembered that the question had quite a large blank space and I thought it was a 4 marker. I could be wrong though, what other parts to that question was there? I might be thinking of the wrong question
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    (Original post by js.int)
    I don't know tbh... I remembered that the question had quite a large blank space and I thought it was a 4 marker. I could be wrong though, what other parts to that question was there? I might be thinking of the wrong question
    The question just asked to find the specific charge. There likely was a large space for the derivations as they were expecting them but they'll also surely be expecting people to just have learnt the formula... and account for mistakes that people may make in remembering it. I'm hoping anyway :P

    I just knew something was wrong, all the questions I've done in the past used figures which gave you the expected charge of 1.8x10^11. Was surprised when it wasn't but didn't get a chance to have a look
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    (Original post by -jordan-)
    The question just asked to find the specific charge. There likely was a large space for the derivations as they were expecting them but they'll also surely be expecting people to just have learnt the formula... and account for mistakes that people may make in remembering it. I'm hoping anyway :P
    Tbh they may put something in the additional guidance if a lot of people are making the mistake. But I think you would still get some marks out of 4.
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    (Original post by js.int)
    I don't know tbh... I remembered that the question had quite a large blank space and I thought it was a 4 marker. I could be wrong though, what other parts to that question was there? I might be thinking of the wrong question
    It was 4 marks because the equation isn't given, so you either have to know the equation or the derivation for it. It asked you to calculate the specific charge given some values of B, r and V - it never asked anything about deriving it. They will probably give marks for derivation, but also the mark scheme will likely say "If 1.9x10^11 Ckg^-1 seen, 4 marks". They can't take marks away because you know the equation MORE than other people, that'd be completely nonsensical.
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    Can anyone remmeber how many marks was in question 5 and how they were split up for each ? I`m a bit panicy as i got the first part wrong and carried my answer onto the rest, will there be marks awarded for error carried forward?

    Thanks
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    (Original post by Jay1421)
    Could you possibly tell me which question this was? I really don't remember it and I'm worried I've missed one
    It was something along the lines of: Give one reason why background radiation has increased within the last 100 years.
 
 
 
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