If the critical frequency (v_(0)) for emission of photoelectrons from a metal is 9.62xx10^(14) s^(-1), then light that can emit photoelectons should have a wavelength equal to

If the critical frequency (v_(0)) for emission of photoelectrons from

1.	If the critical frequency (v_(0)) for emission of photoelectrons from a metal is 9.62xx10^(14) s^(-1), then light that can emit photoelectons should have a wavelength equal to
Answer» 6000Å 5000Å 4500Å 3000Å Solution :The light with lowest `lambda` shall have LARGEST ENERGY. `therefore` it can CAUSE the emission . (In this EQUATION calculation is not required).