The kinetic energies of the photoelectrons are E_(1) and E_(2) with wavelength of incident light lamda_(1) and lamda_(2). The work function of the metal is :

The kinetic energies of the photoelectrons are E_(1) and E_(2) with wa

1.	The kinetic energies of the photoelectrons are E_(1) and E_(2) with wavelength of incident light lamda_(1) and lamda_(2). The work function of the metal is :
Answer» `(E_(1)lamda_(1)-E_(2)lamda_(2))/(lamda_(2)-lamda_(1))` `(E_(1)E_(2))/(lamda_(1)-lamda_(2))` `((E_(1)-E_(2))lamda_(1)lamda_(2))/(lamda_(1)-lamda_(2))` `(lamda_(1)lamda_(2)E_(1))/((lamda_(1)-lamda_(2))E_(2))` Solution :`E=(HC)/(lambda_(1))-W and E_(2)=(hc)/(lambda_(2))-w` `:.(E_(1)+w)/(E_(2))+w=(lambda_(2))/(lambda_(1)) or w=(E_(1) lambda_(1)-E_(2)lambda_(2))/(lambda_(2)-lambda_(1))`

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The kinetic energies of the photoelectrons are E_(1) and E_(2) with wavelength of incident light lamda_(1) and lamda_(2). The work function of the metal is :

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