Please explain and show formula and work. Answer is given already, thank you. Pl

Question

Please explain and show formula and work. Answer is given already, thank you.

Please explain and show formula and work. Answer is given already. thank you. Photons scatter off an electron in a Compton Effect experiment. 1) At what deflection angle relative to its initial path does a photon have the greatest change in wavelength? (where 0 degrees is continuing straight and 180 degrees is deflecting directly back) degrees b 45 degrees 90 degrees d 13 degrees degrees A certain metal in the photoelectric effect experiment has a work function of 4.2 eV. For a given experiment, the maximum kinetic energy of the ejected electrons is 1.8 eV. 2) What is the wavelength of the incident photons? 407 nm 502 nm 3) What is the DeBroglie wavelength of the ejected electrons? 4) What is the maximum length of photo that would eject electrons from this metal? a 205 nm b 393 nm c 495 nm 595 nm e) 69 nm

Explanation / Answer

1)

change in wavelength delta lamda = [h/(mo*c)][1-cos(phi)]..

to get max change in wavelength phi must 180 degrees so that change in wavelength = [(h/mo*c)][1-(-1)] = 2*h/(mo*c)..

so answer is e) 180 degrees..

2) Energy of incident photons E = Kmax + wo = 4.2+1.8 = 6eV = 6*1.6*10^-19 J = 9.6*10^-19 J..
but E = h*c/lamda..
wavelength almda = h*c/E = 6.625*10^-34*3*10^8/9.6*10^-19 = 207 nm..

3) debroglie wavelength lamda = h/p = h/sqrt(2*m*Kmax) = 6.625*10^-34/sqrt(2*9.1*10^-31*1.8*1.6*10^-19) = 0.91 nm...

4) lamda max = (h*c)/wo=6.625*10^-34*3*10^8/(4.2*1.6*10^-19) = 295 nm

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