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What will be the kinetic energy of emitt...

What will be the kinetic energy of emitted photoelectrons if light of threshold frequency falls on a metal?

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The correct Answer is:
zero
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Light of wavelength 6000Å is incident on a metal. To release an electron from the metal surface, 1.77 eV of energy is needed. Find the kinetic energy of the fastest photoelectron. What is the threshold frequency of the metal? (h = 6.62 xx 10^(-27) " erg".s, 1eV = 1.6 xx 10^(-12) " erg")

The work function for cesium is 1.8 eV. A light of wavelength 5000 Å is incident on it. Calculate (i) maximum kinetic energy of emitted electrons, (ii) threshold frequency and threshold wavelength and (iii) maximum velocity of the emitted electrons

Knowledge Check

  • Maximum kinetic energy of photoelectron depends on

    A
    nature of the metal surface
    B
    intensity of the incident radiation
    C
    energy of incident photon
    D
    work function of the metal

  • Einstein's equation for photoelectric effect is E_("max") = hf - W_(0) , where h = Planck's constant = 6.625 xx 10^(-34) J.s, f = frequency of light incident on metal surface, W_(0) = work function of metal and E_("max") = maximum kinetic energy of the emitted photoelectrons. It is evident that if the frequency f is less than a minimum value f_(0) or if the wavelength lamda is greater than a maximum value lamda_(0) , the value of E_("max") would be negative, which is impossible. Thus for a particular metal surface f_(0) is the threshold frequency and lamda_(0) is the threshold wavelength for photoelectric emssion to take place. Again if the collector plate is ketp at a negative potential with respect to the emitter plate, the velocity of the photoelectrons would decrease. The minimum potential for which the velocity of the speediest electron becoes zero, is known as the stopping potential, the photoelectric effect stops for a potential lower than this. [velocity of light = 3xx 10^(8) m.s^(-1) , mass of an electron m = 9.1 xx 10^(-31) kg , charge of an electron, e = 1.6 xx 10^(-19)C Ultraviolet of wavelength 1800 Å is incident on the metal surface. The maximum velocity of the emitted photoelectron (in m.s^(-1) ) is

    A
    `8.5 xx 10^(5)`
    B
    `1.2 xx 10^(6)`
    C
    `1.7 xx 10^(5)`
    D
    `2.4 xx 10^(6)`

  • Einstein's equation for photoelectric effect is E_("max") = hf - W_(0) , where h = Planck's constant = 6.625 xx 10^(-34) J.s, f = frequency of light incident on metal surface, W_(0) = work function of metal and E_("max") = maximum kinetic energy of the emitted photoelectrons. It is evident that if the frequency f is less than a minimum value f_(0) or if the wavelength lamda is greater than a maximum value lamda_(0) , the value of E_("max") would be negative, which is impossible. Thus for a particular metal surface f_(0) is the threshold frequency and lamda_(0) is the threshold wavelength for photoelectric emssion to take place. Again if the collector plate is ketp at a negative potential with respect to the emitter plate, the velocity of the photoelectrons would decrease. The minimum potential for which the velocity of the speediest electron becoes zero, is known as the stopping potential, the photoelectric effect stops for a potential lower than this. [velocity of light = 3xx 10^(8) m.s^(-1) , mass of an electron m = 9.1 xx 10^(-31) kg , charge of an electron, e = 1.6 xx 10^(-19)C The threshold wavelength of photoelectric effect for a metal surface is 4600 Å . Work function of the metal (in eV) is

    A
    `2.7`
    B
    `3.45`
    C
    `4.2`
    D
    `6.9`

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    Light rays of wavelength lambda and lambda/2 are incident on a a photosensitive metal surface, IF the maximum kinetic energy of the emitted photoelectrons from the metal surface in 2nd case be 3 times the maximum kinetic energy of emitted photoelectrons in the 1st case then determine the work function of the metal.

    A photoelectric source is illuminated successively by monochromatic light of wavelenth lamda and (lamda)/(2) . Calculate the work function of material of the source if the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case.

    If a light with frequency 2.0xx10^(16)Hz emitted photoelectrons with double the kinetic energy as are emitted by the light of frequency 1.25xx10^(16)Hz from the same metal surface, calculate the threshold frequency of the metal.

    In case of photoelectric effect, the graph of kinetic energy of photoelectron with respect to the frequency of incident radiation will be a straight line. The slope of this straight line depends on

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