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Work function photoelectric effect question help

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    Hi,
    I have got to an answer which I'm convinced is correct but the MS says otherwise for this question:

    Light of frequency 8.9 x 10^14 Hz is incident upon the surface of a metal. The photoelectrons have a max speed of 6.7 x 10^5 m/s. Calculate the work function of the metal in eV.

    I got to 1.81059... J which comes to 1.13 eV, but the MS says that 2.4 eV is correct.

    Can anyone help?
    Thanks
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    (Original post by george2109)
    Hi,
    I have got to an answer which I'm convinced is correct but the MS says otherwise for this question:

    Light of frequency 8.9 x 10^14 Hz is incident upon the surface of a metal. The photoelectrons have a max speed of 6.7 x 10^5 m/s. Calculate the work function of the metal in eV.

    I got to 1.81059... J which comes to 1.13 eV, but the MS says that 2.4 eV is correct.

    Can anyone help?
    Thanks
    What year was the paper?
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    (Original post by King31)
    What year was the paper?
    It's the new spec specimen paper for edexcel Advanced Physics 2. It's the second one which isn't on their website, my teacher gave it to me.
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    Can you post your workings please
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    l got 2.4 eV...U know kinetic energy=1/2 mv^2 , mass of an electron is 9.11 times 10^-31. U should get a kinetic enegy value of 2.0441 times 10^-19 and work out from there, l hope this helps
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    Have managed to get the answer the mark scheme is asking for
    6.63x10^-34 x 8.9x10^14 = work function + max kinetic energy


    Where max kinetic energy is 0.5mv^2 so Ekmax = 0.5 x9.11x10^31 x v^2
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    (Original post by Olly Nourse)
    Have managed to get the answer the mark scheme is asking for
    6.63x10^-34 x 8.9x10^14 = work function + max kinetic energy


    Where max kinetic energy is 0.5mv^2 so Ekmax = 0.5 x9.11x10^31 x v^2
    Yes, thanks alot! I was being stupid - I got the k.e. value to be 4 x 10^-19 not 2 x 10^-19
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    (Original post by King31)
    l got 2.4 eV...U know kinetic energy=1/2 mv^2 , mass of an electron is 9.11 times 10^-31. U should get a kinetic enegy value of 2.0441 times 10^-19 and work out from there, l hope this helps
    Yes, thank you. My k.e. was wrong but I've got to the right answer now!
 
 
 
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