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    Could some one please help me out, i can't work out how to do it
    Any help would be great.
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    (Original post by droid)
    Could some one please help me out, i can't work out how to do it
    Any help would be great.
    what syllabus u doin mate?
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    Edexcel GCE
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    (Original post by droid)
    Edexcel GCE
    ah rite, im not doin that, but if u typed out the question id be happy to have a go at it. there seems to be a lot of sylabus overlap
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    a) Intensity = power/area(4[pi]r^2)

    b) the table:

    Well we know energy of one photon = hc/landa, so substituting the two beams into it, and finding the ratio of Ea/Eb gives us 1:1.5 (or 2/3 in fraction).

    Now this part requires more thought. N is the Number of photons. They have said that they both have the SAME intensity. We have just found out that beam A has 150% more energy per photon than beam B. So for the two beams to be of the same intensity, one of them has to have a great number of photons. Meaning that Beam B must have more photons than Beam A, to make them equal the same intensity.
    So the ratio of the number of Na/Nb must be 1.5:1 which is 3/2 .

    C) Work function is the minimum energy required to liberate an electron for the surface of a metal.

    The photon energies in each beam:
    Beam A = 6.67x10^-19 OR 4.17eV
    Beam B = 4.45x10^-19 OR 2.78eV

    So looking at the energies, we can see that Beam B can only liberate electrons from Potassium.
    And Beam A can liberate electrons from Potassium and Magnesium.

    The question asks which metal does ONLY beam A emmite electrons from, which is magenesium.
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    (Original post by SinghFello)
    a) Intensity = power/area(4[pi]r^2)

    b) the table:

    Well we know energy of one photon = hc/landa, so substituting the two beams into it, and finding the ratio of Ea/Eb gives us 1:1.5 (or 2/3 in fraction).

    Now this part requires more thought. N is the Number of photons. They have said that they both have the SAME intensity. We have just found out that beam A has 150% more energy per photon than beam B. So for the two beams to be of the same intensity, one of them has to have a great number of photons. Meaning that Beam B must have more photons than Beam A, to make them equal the same intensity.
    So the ratio of the number of Na/Nb must be 1.5:1 which is 3/2 .

    C) Work function is the minimum energy required to liberate an electron for the surface of a metal.

    The photon energies in each beam:
    Beam A = 6.67x10^-19 OR 4.17eV
    Beam B = 4.45x10^-19 OR 2.78eV

    So looking at the energies, we can see that Beam B can only liberate electrons from Potassium.
    And Beam A can liberate electrons from Potassium and Magnesium.

    The question asks which metal does ONLY beam A emmite electrons from, which is magenesium.
    I get the same except for in part b) I get the ratios the other way round. i.e. I get the first one to be 3/2 and the second to be 2/3 ... :confused:
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    (Original post by sonja)
    I get the same except for in part b) I get the ratios the other way round. i.e. I get the first one to be 3/2 and the second to be 2/3 ... :confused:
    you are correct
 
 
 
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