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Electrons are emitted with zero velocity...

Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 `overset@A`. Calculate the threshold frequency `(v_0)` and work function `(w_0)` of the metal.

Answer

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Knowledge Check

  • The kinetic energy of the most energetic photoelectrons emitted from a metal surface is doubled when the wavelength of the incident radiation is reduced from lamda_1 to lamda_2 The work function of the metal is:

    A
    `(hc)/(lamda_1lamda_2)(2lamda_2-lamda_1)`
    B
    `(hc)/(lamda_1lamda_2)(2lamda_1-lamda_2)`
    C
    `(hc)/(lamda_1lamda_2)(lamda_1+lamda_2)`
    D
    `(hc)/(lamda_1lamda_2)(lamda_1-lamda_2)`

  • The kinetic energy of the most energetic photoelectrons emitted from a metal surface is doubled when the wavelength of the incident radiation is reduced from lamda_(1) to lamda_(2). The work function of the metal is

    A
    `(hc)/( lamda _(1) lamda _(2)) ( 2 lamda _(2) - lamda _(1))`
    B
    `(hc)/( lamda _(1) lamda _(2)) ( 2 lamda _(1) - lamda _(2))`
    C
    `(hc)/( lamda _(1) lamda _(2)) ( 2 lamda _(1) + lamda _(2))`
    D
    `(hc)/( lamda _(1) lamda _(2)) ( 2 lamda _(1) -lamda _(2))`

  • Photons of frequency, v, fall on metal surface for which the threshold frequency is v_(0) . Then

    A
    All ejected electrons have the same kinetic energy, `h(v-v_(0)`):
    B
    The ejected electrons have a distribution of kinetic energy from zero to `h(v = v_(0)`)
    C
    The most energetic electron has kinetic energy hv.
    D
    The average kinetic energy of ejected electrons is `h(v-v_(0)`).

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    To emit a free electron from a metal surface a minimum amount of energy must be supplied.When light of frequency 7.21 * 10^14 Hz is incident on a metal surface, the maximum speed of the ejected electrons is 6 x 10^5 m/s . Calculate the threshold frequency for the metal.

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