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

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

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

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

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

Text Solution

Verified by Experts

The correct Answer is:

`(hc)/lambda_(1)=W+E_(1),(hc)/lambda_(2)=W+E_(2)`

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NARAYNA-DUAL NATURE-LEVEL-II (C.W)

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