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In photoelectric emission from a metal, ...

In photoelectric emission from a metal, work function `phi_(0)` and threshold frequency `v_(0)` are related as _____.

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`phi_(0)=hv_(0)`
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Knowledge Check

  • A photon of frequency v causes photo electric emission from a surface with threshold frequency v_(0) . The de-Broglies wavelength (lambda) of the photo electron emitted is given by:

    A
    `Deltan=(h)/(2mlambda)`
    B
    `Deltan=(h)/(lambda)`
    C
    `[1/v_(0)-1/v]=(mc^(2))/(h)`
    D
    `lambda=sqrt(h/(2mDeltaV))`

  • If the photoelectric work function of a metal is 5 eV , the threshold frequency of the metal is

    A
    `2.2 xx 10^(15)` Hz
    B
    `1.2 xx 10^(15)` Hz
    C
    `1.5 xx 10^(15)` Hz
    D
    `5.1 xx 10^(15)` Hz

  • Photoelectric emission is observed from a metal surface with incident frequencies v_(1) and v_(2) where v_(1) gt v_(2) . If the kinetic energies of the photoelectrons emitted in the two cases are in the ratio 2 : 1 , then the threshold frequency v_(0) of the metal is

    A
    `v_(1) = v_(2)`
    B
    `(v_(1) - v_(2))/(h)`
    C
    `2 v_(1) - v_(2)`
    D
    `2v_(2) - v_(1)`

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    Photoelectric emission is observed from a metal surface for frequencies v_(1) and v_(2) incidents radiations (v_(1) gt v_(2)) . If the maximum kinetic energy of the two electrons are in the ratio 1:2 , then how is threshold frequency (v^(@)) experssed.

    The photoelectric emission requires a threshold frequency v_0 . For a certian metal lambda _1 = 2200 Å and lambda _2 = 1900 Å produce electrons with a maximum kinetic enrgy KE_1 and KE_2 , if KE_2= 2KE_1 , calculate v_0 and corresponding lambda _0 ,

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    For photoelectric emission from certain metal the cut - off frequency is v . If radiation of frequency 2 v impinges on the metal plate , the maximum possible velocity of the emitted electron will be ( m is the electron mass).

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