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urgent phys query(diffraction grating experiment) watch

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    hi,

    need help

    question asks to describe how you would observe the emission spectrum of hydrogen in the lab.

    markscheme says to view exited hydrogen through difffraction grating and observe emmision spectra

    i said pass high voltage through H gas in a container and pass white light through it, and view through other side with diffraction grating. said that absorpsion spectrum would be viewwed and that 4 dark lines would be visable

    woudl i get marks for this alternative method - its basically the opposite, but not on the markscheme i've found!!!!

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    whats the method on the markscheme?
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    (Original post by Phil23)
    hi,



    markscheme says to view exited hydrogen through difffraction grating and observe emmision spectra

    i said pass high voltage through H gas in a container and pass white light through it, and view through other side with diffraction grating. said that absorpsion spectrum would be viewwed and that 4 dark lines would be visable

    Umm, u'll probs get some marks. The bit about voltage is correct, then view thru a diffraction grating u will see couloured lines on a dark background...from looking at loadsa mark schemes i have seen u shud state the observations as well.
    Now my turn to ask a question...whats a stopping potential, what questions would u get? :confused:
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    (Original post by Phil23)
    hi,

    need help

    question asks to describe how you would observe the emission spectrum of hydrogen in the lab.

    markscheme says to view exited hydrogen through difffraction grating and observe emmision spectra

    i said pass high voltage through H gas in a container and pass white light through it, and view through other side with diffraction grating. said that absorpsion spectrum would be viewwed and that 4 dark lines would be visable

    woudl i get marks for this alternative method - its basically the opposite, but not on the markscheme i've found!!!!

    Isn't your method pretty much the same? you are exiciting the hydrogen atoms with electricity, then observing it through a diffraction grating. As far as i can see, you've got it smack on.

    EDIT: aha - difference between absorption and emmission spectrum - you get an emmission spectrum when you excite atoms, so you wouldn't want to pass white light through it because that would fill in the black gaps in the spectrum. In an emmission spectrum you DONT pass white light through it - you jsut observe the light given out. So your method is getting a little confused. Almost certainly 2/4 marks. Probably 2/3. Either that or I'm getting confused
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    (Original post by Phil23)
    hi,

    need help

    question asks to describe how you would observe the emission spectrum of hydrogen in the lab.

    markscheme says to view exited hydrogen through difffraction grating and observe emmision spectra

    i said pass high voltage through H gas in a container and pass white light through it, and view through other side with diffraction grating. said that absorpsion spectrum would be viewwed and that 4 dark lines would be visable

    woudl i get marks for this alternative method - its basically the opposite, but not on the markscheme i've found!!!!

    Looks reasonably ok to me - although lacking in a little detail if you want many marks for it. Does the mark scheme say spectrometer?
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    It depends who is marking your paper. They are supposed to give you marks if you know what your taking about. But I found when I did PHY4 in Jan, they usually dont.

    Out of curiosity, why is the large voltage needed in the gas?
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    (Original post by scaredashell)
    Umm, u'll probs get some marks. The bit about voltage is correct, then view thru a diffraction grating u will see couloured lines on a dark background...from looking at loadsa mark schemes i have seen u shud state the observations as well.
    Now my turn to ask a question...whats a stopping potential, what questions would u get? :confused:
    Stoppimg potential? As in photo elecric effect?

    The voltage required to reduce the current to zero. Usually between 2 plates where you havwe a flow of electrons. You then place a negative stopping potential on the plate so that the electrons cant reach it, hence reducing the current to zero.
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    (Original post by scaredashell)
    Umm, u'll probs get some marks. The bit about voltage is correct, then view thru a diffraction grating u will see couloured lines on a dark background...from looking at loadsa mark schemes i have seen u shud state the observations as well.
    Now my turn to ask a question...whats a stopping potential, what questions would u get? :confused:

    Stopping potential:

    qVs= hf - phi

    you should know how to rearrange this into y-mx+c
    which is Vs = h/q f - phi/q

    So plotting a graph of Vs against f, gives a gradient equaling h/q, and intercept of phi/q.
    You should know this experiement.

    And also qVs=1/2mv^2
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    Could i just get back to the initial question for a sec - isn't the emmission spectrum only visible when no other (visible) light is present? So the only thing wrong with Phil23's answer is that he passed white light through it? Because an emmission spectrum is light given out...he excited the hydrogen with electricity, so the excited electrons should drop back to ground state, emmitting photons of light.
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    (Original post by CrazyChemist)
    Isn't your method pretty much the same? you are exiciting the hydrogen atoms with electricity, then observing it through a diffraction grating. As far as i can see, you've got it smack on.

    EDIT: aha - difference between absorption and emmission spectrum - you get an emmission spectrum when you excite atoms, so you wouldn't want to pass white light through it because that would fill in the black gaps in the spectrum. In an emmission spectrum you DONT pass white light through it - you jsut observe the light given out. So your method is getting a little confused. Almost certainly 2/4 marks. Probably 2/3. Either that or I'm getting confused

    i think my method is right...did it for synoptic - well what happens is that if you pass white light through exited hydrogen gas then the light from the white one is absorbed and scattered in all directions, so when you view it through a diffraction you see 4 dark lines which correspond to hydrogen gas, instead of 4 bright lines - the rest of the spectrum would not be affected form what i gather

    if that is right then i got 87% on the jan 2003 paper - woo hoo :cool:
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    (Original post by CrazyChemist)
    Could i just get back to the initial question for a sec - isn't the emmission spectrum only visible when no other (visible) light is present? So the only thing wrong with Phil23's answer is that he passed white light through it? Because an emmission spectrum is light given out...he excited the hydrogen with electricity, so the excited electrons should drop back to ground state, emmitting photons of light.
    bingo! hence lines corresponding to H is scattered in all directions, hence the black lines duw to low intensity of this light etc
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    (Original post by F1 fanatic)
    Stoppimg potential? As in photo elecric effect?

    The voltage required to reduce the current to zero. Usually between 2 plates where you havwe a flow of electrons. You then place a negative stopping potential on the plate so that the electrons cant reach it, hence reducing the current to zero.
    Thanks!!!! I think i get it!
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    aha...i see.
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    (Original post by SinghFello)
    Stopping potential:

    qVs= hf - phi

    you should know how to rearrange this into y-mx+c
    which is Vs = h/q f - phi/q

    So plotting a graph of Vs against f, gives a gradient equaling h/q, and intercept of phi/q.
    You should know this experiement.

    And also qVs=1/2mv^2
    Got it! lol. I was wondering y there was a graph involved in one of the questions i looked at. cheers
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    u're on d right track, when the Hydrogen gas is excited, they gain energy. Now, when the return back to their groundstate(or initial position) they release this excess enrgy gained in form of PHOTONS(light). This is the light that is observed as emission spectra. This emission been distinctive to individual elements. So, there shouldn't be any light(visible) presesnt, as u know it is observed(spectrum) in a dark background.
    Hope this helps
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    hmm! still don't quite get d reason why u are observing the black lines(absorption spectrum, right?). But are u sure of this method, although it looks very genius!
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    (Original post by mikeA1)
    hmm! still don't quite get d reason why u are observing the black lines(absorption spectrum, right?). But are u sure of this method, although it looks very genius!
    pretty sure - i learnt about teh emission spectra after the absorption one, lol; from what i understand is that the absortion and emission spectra give the opposte results - i.e. dark lines where H should occur, opposed to the bright lines on a dark background, as is teh case in an emission spectra

    nothing genius abour it - its in phy 6 i think
 
 
 
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