Question

Photons with a frequency of 8.9 1015 Hz shine on a piece of platinum (work function = 6.4 eV). What is the energy of the ejected electrons?

What is the de Broglie wavelength of the ejected electrons?

Answer 1

energy of incident photon E = h*f = 6.626*10^-34*8.9*10^15 = 5.89*10^-18 J

work function Wo = 6.4 eV = 6.4*1.6*10^-19 = 1.024*10^-18 J

from photo electric equation

E = Wo + KE

energy of emitted electrons KE = E - Wo = 5.89*10^-18 - 1.024*10^-18 = 4.866*10^-18 J

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KE , K= (1/2)*m*^2

speed v = sqrt(2*K*m)

De brogile wavelength lambda = h/mv = h/sqrt(2*K*m)

lambda = 6.626*10^-34/(sqrt(2*4.866*10^-8*9.11*10^-31) = 2.22*10^-15 m