Light with a wavelength of 425 nm falls on a photoelectric surface that has a work function of 2.00 eV. What is the maximum speed of any emitted photoelectrons?
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λ= 425 nm = 425 x 10-9 m . w = 2 eV = 3.2043 x 10-19 J . KE of light Ek = Ep - w Ek⇒ [ (hc)/λ ] - w . c... View the full answer