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    I'm doing a question which says to calculate the de broglie wavelength of a proton at speed v.
    I have to use the formula -

    ? = h / (2MeV)^0.5 e - charge of an electron, M - mass, V - accelerating voltage

    That's quite badly typed out, i'm afraid.
    Should i be using the mass of a proton or electron?

    Thank youuu
    -Afsaneh-
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    (Original post by Churchgirl)
    I'm doing a question which says to calculate the de broglie wavelength of a proton at speed v.
    I have to use the formula -

    ? = h / (2MeV)^0.5 e - charge of an electron, M - mass, V - accelerating voltage

    That's quite badly typed out, i'm afraid.
    Should i be using the mass of a proton or electron?

    Thank youuu
    -Afsaneh-
    Mass of a proton, definitely. The de broglie wavelength of a proton is Planck's constant over momentum of the proton, and momentum of a proton has nothing to do with the mass of an electron That's right that e in your formula is the charge of an electron, but it's the charge of a proton as well.
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    (Original post by jaroc)
    Mass of a proton, definitely. The de broglie wavelength of a proton is Planck's constant over momentum of the proton, and momentum of a proton has nothing to do with the mass of an electron That's right that e in your formula is the charge of an electron, but it's the charge of a proton as well.
    I understand, thankfully

    Can i ask you for help on one more part, it's just i'm not too sure with special relativity.

    Question -
    Protons are accelerated from rest to a speed, v through a pd of 2900MV.
    Show that the mass of a proton at this speed is 4.1×rest mass of a proton.
    Answer -
    I did eV = Kinetic energy
    I found the speed to be 7.45×10^8
    But when i substitute it into the special rel equation for mass, i get a -ve value in the square root.

    Do you know what to do?
    Much apprecited!

    PS - Congratulations on your university offers!
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    (Original post by Churchgirl)
    I understand, thankfully

    Can i ask you for help on one more part, it's just i'm not too sure with special relativity.

    Question -
    Protons are accelerated from rest to a speed, v through a pd of 2900MV.
    Show that the mass of a proton at this speed is 4.1×rest mass of a proton.
    Answer -
    I did eV = Kinetic energy
    I found the speed to be 7.45×10^8
    But when i substitute it into the special rel equation for mass, i get a -ve value in the square root.

    Do you know what to do?
    Much apprecited!

    PS - Congratulations on your university offers!
     m = \gamma m_0

    Find v, use that to find gamma and you're done.
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    (Original post by Prime Suspect)
     m = \gamma m_0

    Find v, use that to find gamma and you're done.

    I'm not too sure if i've even come across that formula before O.o
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    (Original post by Churchgirl)
    I'm not too sure if i've even come across that formula before O.o
    Derivation and some other stuff here

    Something's wrong in your case though as the velocity you've got is greater than c?
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    (Original post by Prime Suspect)
    Derivation and some other stuff here

    Something's wrong in your case though as the velocity you've got is greater than c?
    All the forrmulae i can use are in:
    http://store.aqa.org.uk/qual/gceasa/...W-QP-JUN07.PDF
    On page 4
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    (Original post by Churchgirl)
    All the forrmulae i can use are in:
    http://store.aqa.org.uk/qual/gceasa/...W-QP-JUN07.PDF
    On page 4
    That formula is on page 4, just written as:

     E = mc^2 = \dfrac{m_0c^2}{(1-\frac{v^2}{c^2})^{\frac{1}{2}}}

    I think though that the problem is that you've got a velocity greater than c - have you checked your numbers?
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    Well to find v -
    eV = Kinetic energy
    Rearranged :
    v = (2eV / m)^0.5
    2(1.6× 10^-19)(2900^6)/(1.67× 10^-27) then square rooted the whole thing
    I get : 7.45 10^8

    I can't see where my mistake is but i'm sure i've made one
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    (Original post by Churchgirl)
    Well to find v -
    eV = Kinetic energy
    Rearranged :
    v = (2eV / m)^0.5
    2(1.6× 10^-19)(2900^6)/(1.67× 10^-27) then square rooted the whole thing
    I get : 7.45 10^8

    I can't see where my mistake is but i'm sure i've made one
    No that is correct; I think I know what the problem is though - you're working in the lab frame (as opposed to the particle's rest frame) when making this calculation, so you can't use the rest mass of the proton in this equation when you find kinetic energy

    I think you need to say that

     E = m_0c^2 + eV

    i.e. the total energy E is equal to the sum of the rest mass energy m_0c^2 and the kinetic energy (which we know is equal to eV).

    You then use that the total energy E is also given by

     E = \gamma m_0 c^2 from your formula book, so putting together:

     \gamma m_0 c^2 = m_0c^2 + eV

    i.e.

     \gamma = 1 + \dfrac{eV}{m_0c^2}

    Work this out and I think you should have your answer
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    (Original post by Prime Suspect)
    No that is correct; I think I know what the problem is though - you're working in the lab frame (as opposed to the particle's rest frame) when making this calculation, so you can't use the rest mass of the proton in this equation when you find kinetic energy

    I think you need to say that

     E = m_0c^2 + eV

    i.e. the total energy E is equal to the sum of the rest mass energy m_0c^2 and the kinetic energy (which we know is equal to eV).

    You then use that the total energy E is also given by

     E = \gamma m_0 c^2 from your formula book, so putting together:

     \gamma m_0 c^2 = m_0c^2 + eV

    i.e.

     \gamma = 1 + \dfrac{eV}{m_0c^2}

    Work this out and I think you should have your answer

    Needless to say, i wouldn't have thought of that. Especially for a 3 mark question!
    Thank you so much . . . Prime!
    I'll positive rep you
 
 
 
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