Physics

The work function for potassium and cesium are 2.25 and 2.14 eV, respectively.
a) Will the photoelectric effect occur for either of these elements with incident light of wavelength 565 nm?
b) With light of wavelength 518 nm ?

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  1. convert 565 and 518 to wavelength.

    Fnergy= hc/lambda

    set lambda=hc/energy

    now convert eV to joules, plug it in and compute lambda cutoff for each metal. If lambda oncoming is shorter than 565, and 518nm, photoelectric effect occurs.

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    bobpursley
  2. W potassium eV 0 ( ) = 2.25 ,
    W caesium eV 0 ( ) = 2.14 ,
    7
    1 5650 10 m

    λ = × ,
    7
    2 5180 10 m

    λ = ×
    34
    h J s 6.63 10 .

    = × ,
    8
    c m s = × 3 10 /
    W potassium eV 0 ( ) = 2.25
    19
    2.25 1.6 10 J

    = × ×
    19
    3.6 10 J

    = ×
    W caesium eV 0 ( ) = 2.14
    19
    2.14 1.6 10 J

    = × ×
    19
    3.424 10 J

    = ×
    i.
    34 8
    7
    1
    6.63 10 3 10
    5.650 10
    hc


    × × ×
    =
    � ×

    19
    3.52 10 J

    = ×
    This energy (of the
    incident photon) is greater
    than W0
    (caesium), but
    less than W0
    (potassium).
    Hence, photoelectric effect
    will occur in case of
    caesium, but not in case
    of potassium.
    ii.
    34 8
    7
    2
    6.63 10 3 10
    5.180 10
    hc


    × × ×
    =
    λ ×

    19
    3.84 10 J

    = ×
    This is greater than W0
    for potassium and for
    caesium. Hence,
    photoelectric effect will occur in both cases.

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