What would the minimum work function for a metal have to be for visible light (h

Question

What would the minimum work function for a metal have to be for visible light (having wavelengths between 400 and 700 nm ) to eject photoelectrons?

Explanation / Answer

photoelectric equ is hc/? = phi + 0.5 mv^2 incident energy of photon = work function + KE of electrons --------------------------------------… the work function of any metal is Fixed, and whatever extra energy the photons entail is imparted to electrons as their kinetic energy. If photon is short of the work function energy of metal, no electrons will be ejected from within metal. ----------------------------- the above question implies >>> assume photo electrons are ejected with 400-700 nm range in visible light?? assume >> they are just liberated so KE = o ---------------------------- hc/? = phi (Phi)min = hc / ?max (in Range given) so ?max = 700 nm will make the phi minimum (Phi)min = 6.63*10^-34*3*10^8/ 700*10^-9 Joule (Phi)min = 2.828*10^-19 Joule (Phi)min = 2.84*10^-19 / 1.6*10^-19 = 1.7755 eV --------------------------------------… 400nm photons will be having 3.1071 eV energy, which will be much above the minimum criterion, and will impart KE to electrons as well Source(s): one needs to read this in totality, and along with latent/ expressed constraints

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