Light of wavelength 450 nm is shone on a piece of metal; the stopping potential

Question

Light of wavelength 450 nm is shone on a piece of metal; the stopping potential for the photoelectrons is measured at 0.52 V. a. Determine the work function of the metal. b. If the area of the metal shone is 1 cm^2, and the effective intensity of the light used is 10^-3 W/m^2, how many photoelectron will be knocked out from the metal? Assume 100% efficiency, i.e. one photon will knock-off one electron. c. According to classical physics, a photoelectron will be released if the atom absorbs energy of at least equal to the work function of the material (obtained in part a). Assuming the effective area of an atom exposed to the beam is 3.2 times 10^-20 m^2, how long do we have to wait since we turn on the beam until we observe the emission of photoelectrons if classical physics were true?

Explanation / Answer

given

lamda = 450 nm

Vo = 0.52 V

a)
q*Vo = E - Wo

==> Wo = E - q*Vo

= h*c/lamda - q*Vo

= 6.626*10^-34*3*10^8/(450*10^-9) - 1.6*10^-19*0.52

= 3.58*10^-19 J or 2.24 eV

b)

we know, Intensity = Power/Area

Power = Intensity*Area

= 10^-3*10^-4

= 10^-7 W

no of photo electrons emitted per second, N = P/E_photon

= P/(h*c/lamda)

= 10^-7/(6.626*10^-34*3*10^8/(450*10^-9))

= 2.26*10^11

c) Energy obsorbed per unit time, E = 10^-3*3.2*10^-20

= 3.2*10^-23 J

so, time taken = Workfunction/Energy oborbed in one second

= 3.58*10^-19/(3.2*10^-23)

= 11187 s or 186 minutes or 3.1 hours

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