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Thread: Photoelectric Effect

  1. #1 Photoelectric Effect 
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    I was wondering if anyone could help me with this question....

    What is the maximum number of electrons that can be emitted if a potassium surface of work function 2.40 eV absorbs 3.25 x 10^-19 J of radiation at a wavelength of 300 nm? What is the kinetic energy and velocity of the electrons emitted?

    I really have no idea so any help would be much appreciated.

    Thanks


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  3. #2  
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    What is the maximum number of electrons that can be emitted if a potassium surface of work function 2.40 eV absorbs 3.25 x 10^-19 J of radiation at a wavelength of 300 nm? What is the kinetic energy and velocity of the electrons emitted?

    3.35 x 10<sup>-19</sup> J appears to be less than the energy of one photon at a wavelength of 300 nm and is also less than the work function !

    (1 eV = 1.6 x 10<sup>-19</sup> J so 2.40 eV = 3.84 x 10<sup>-19</sup> J.)

    No electrons would be emitted. Are you sure that you have got the numbers right?


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  4. #3  
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    OH no sorry I dont have the numbers right......instead of 3.25x10^-19 it is 3.25x10^-3

    Thanks for mentioning it
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  5. #4  
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    .....it is 3.25x10^-3

    In that case, to find the maximum number of electrons emitted, you first recognise that the absorption of one photon leads, at most, to the emission of one electron. So you need to find out how many photons are required to give a total energy of 3.25 x 10<sup>-3</sup> J. This is done simply by dividing the energy of one photon into 3.25 x 10<sup>-3</sup> J.

    The energy of a single photon at frequency v is hv, in which h is Planck's constant. To find out what v is for light of wavelength 300 nm, use:

    speed of light = v x wavelength.

    In other words, the energy of one photon is hc/wavelength.

    I get a value of 6.26 x 10<sup>-19</sup> J. Dividing this into 3.25 x 10<sup>-3</sup> gives 5.19 x 10<sup>15</sup>, which is the maximum number of electrons that can be emitted.

    6.26 x 10<sup>-19</sup> J is the same as 3.91 eV ( 1 eV = 1.6 x 10<sup>-19</sup> J). So the energy of the emitted electrons will be this minus the value of the work function, i.e. (3.91 - 2.40) eV. You will be able to work out that this comes to something like 2.41 x 10<sup>-19</sup> J. This will be the maximum kinetic energy of each emitted electron. To find other things, such as the maximum velocity of an emitted electron, use:

    kinetic energy = 1/2mv<sup>2</sup>,


    Make sure that you express quantities in the correct units (e.g. 300 nm is 300 x 10<sup>-9</sup> m), and please check the numbers given above yourself.
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  6. #5  
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    Thanks so much that was of great help
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