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A monochromatic light soure of frequency...

A monochromatic light soure of frequency f illuminates a metallic surface and ejects photoelectrons. The photoelectrons having maximum energy are just able to ionize the hydrogen atoms in ground state. When the whole experiment is repeated with an incident radiation of frequency `(5)/(6) f`, the photoelectrons so emitted are able to excite the hydrogen atom beam which then emits a radiation of wavelength `1215 Å`. (a) What is the frequency of radiation? (b) Find the work- function of the metal.

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D
(a) Using Einstein's equation of photoelectric effect, `K_(max) = hf -w`
here, `K_(max) = 13.6 eV`
`:. Hf - w =13.6 eV` …(i)
Further, h((5)/(6)f) - W = (12375)/(1215) =10.2 eV` (ii)
Solving Eqs. (i) and (ii) we have ` (hf)/(6) = 3.4 eV or f = ((6)(3.4)(1.6)xx10^(-19))/(6.63xx10^(-34))`
`= 4.92xx10^15 Hz`
(b) W = hf - 13.6`
`=6(3.4)-13.6 = 6.8 eV`
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Knowledge Check

  • When light is incident on a surface, photoelectrons are emitted. For photoelectrons,

    A
    the value of kinetic energy is same.
    B
    kinetic energy does not depend on the wavelength of incident light.
    C
    the value of kinetic energy is more than a certain minimum energy.
    D
    none of the above.

  • When the frequency of the incident radiation on a metallic plate is doubled, KE of the photoelectrons will be

    A
    doubled
    B
    halved
    C
    more than doubled
    D
    increases but less than doubled.

  • When the frequency of incident radiation on a metallic plate is doubled, KE of photoelectron will be

    A
    doubled
    B
    halved
    C
    more than double
    D
    increases but less than double.

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