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If light of wavelength lambda(1) is allo...

If light of wavelength `lambda_(1)` is allowed to fall on a metal , then kinetic energy of photoelectrons emitted is `E_(1)`. If wavelength of light changes to `lambda_(2)` then kinetic energy of electrons changes to `E_(2)`. Then work function of the metal is

A

`(E_(2)lambda_(1)-E_(1)lambda_(2))/lambda_(1)`

B

`(E_(1)lambda_(1)-E_(2)lambda_(2))/(lambda_(1)+lambda_(2))`

C

`(E_(1)lambda_(1)+E_(2)lambda_(2))/(lambda_(1)-lambda_(2))`

D

`(E_(2)lambda_(2)-E_(1)lambda_(1))/(lambda_(1)-lambda_(2))`

Text Solution

Verified by Experts

The correct Answer is:
D
`(hc)/lambda_(1)=W+E_(1),(hc)/lambda_(2)=W+E_(2)`
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Knowledge Check

  • Maximum kinetic energy of a photoelectron is E when the wavelength of incident light is lambda . If energy becomes four times when wavelength is reduced to one thrid, then work function of the metal is

    A
    `(3hc)/(lambda)`
    B
    `(hc)/(3lambda)`
    C
    `(hc)/(lambda)`
    D
    `(hc)/(2lambda)`

  • In a photo-emissive cell, with exciting wavelength lambda_1 , the maximum kinetic energy of electron is K. If the exciting wavelength is changed to lambda_2 , the maximum kinetic energy of electron is 2K then

    A
    `lambda_1=2lambda_2`
    B
    `lambda_1 gt 2lambda_2`
    C
    `lambda_1 lt 2lambda_2`
    D
    `lambda_1=lambda_2/2`

  • The kinetic energy of most energetic electrons emitted from a metallic surface is doubled when the wavelength lamda of the incident radiation is changed from 400 nm to 310 nm. The work function of the metal is

    A
    0.9 eV
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    D
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