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The photoelectric threshold frequency fo...

The photoelectric threshold frequency for potassium is `3 xx 10^(14)` Hz. The work function for potassium is `(h=6.625 xx 10^(-34)Js)`

A

50 nm

B

60 nm

C

500 nm

D

600 nm

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A
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Knowledge Check

  • Frequency of a wave is 6 xx 10^(15) Hz. The wave is

    A
    Radiowave
    B
    Microwaves
    C
    X-ray
    D
    None of these.

  • The threshold wavelength for the ejection of electron for metal X is 330 nm. The work function for photoelectric emission from metal X is (h=6.6xx10^(-34)Js)

    A
    `1.2xx10^(-18)J`
    B
    `1.2xx10^(20)J`
    C
    `6xx10^(-19)J`
    D
    `6xx10^(-12)J`

  • The momentum of a photon of an electromagnetic radiation is 3.3xx10^(-29) kg m/s. The frequcncy of the associated wave is (h=6.6xx10^(-34)Js,c=3xx10^(8)m//s)

    A
    `3.0xx10^(3)Hz`
    B
    `6.0xx10^(3)Hz`
    C
    `7.5xx10^(12)Hz`
    D
    `1.5xx10^(13)Hz`.

    • Similar Questions

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    The threshold frequency for a certain metal is 3.3 xx 10^(14) Hz . If light of frequency 8.2 xx 10^(14) Hz is incident on the metal, predict the cutoff voltage for the photoelectric emission.

    When light of frequency 6 xx 10^(14) Hz is incident on a photosensitive metal, the kinetic energy of the photoelectron ejected is 2e V. Calculate the kinetic energy of the photoelectron in eV when light to frequency 5 xx 10^(14) Hz is incident on the same metal. Given Planck's constant, h = 6.625 xx 10^(-34) Js , 1 eV = 1.6 xx 10^(-19) .

    A photon of energy 10 eV is incident on a metal surface of threshold frequency 1.6 xx 10^15 Hz . The kinetic energy of the photoelectron (in eV) is (assume h = 6 x 10^-34 )

    A photon of energy 8 eV is incident on a metal surface of threshold frequency 1.6 xx 10^15 Hz . The K.E. of the photoelectrons emitted (in eV) is (Take h = 6 xx 10^-34 J - s )

    A photon of energy 8 eV is incident on a metal surface of threshold frequency 1.6 xx 10^15 Hz . The K.E. of the photoelectrons emitted is (in eV). Take h as 6 xx 10^-34 J s .

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