Practice Problem 28.1 SOLUTION c/f Combining these SET UP The ener y Ro1a photon

Question

Practice Problem 28.1 SOLUTION c/f Combining these SET UP The ener y Ro1a photon related to its frequan y f by E -h and 20hing for , we get = hc/E length is related to the requency by the general wave relationship Let us explore the relationship between the energy snd wavelength of photon. Silicon ims become better electrical conductors when they are illuminated by phatons with energiss of 1.14 eV or graater. This bahavior ia called photooonduatty. What is the coresponding photon wawelangth? In what portian of the electromagnetic spectrum does it lie? Tha a lationships SOLVE It's easiest to perform our calculations using units of eV-s) for that is h-4 136 x 10 15 eV s Usng the preceding expression for we get he= (4.136×10-ss eve)(3.00x1(f myla) = 1.00 x 10-6 m = 1000 nm The wavelengths of the visible spectrum are about 400 to 700 nm, so the wavelength we have found is in the near-infrered region of the spectrum. REFLECT The frequency of a photon is inversely proportional to its wavelength, so the minimum energy of 1.14 eV corresponds to the maximum wavelength that couses photoconduotion for silicon in this oase, about 1000 nm. Thus, for silicon, all ight with a wavclength shorter than 100 nm, indluding al light in the visibic pectrum, contributes to photoconductivity Part A - Practice Problem: If we n ed material that is photocondu Ne for any wavelength in the visible spectrum (La., 400 to 700 nm). to what range of photon nergies must it respond? Express your answers in clectron-volts to three signiticant tigures, separated by a comma. Submit My Answers Give Up

Explanation / Answer

wavelength and Eneergy are related by E = hc/L

so for wavelength L = 400 nm

Emax = (6.626 e-34*3e8)/(400 e -9)

Emax = 3.105 e V

for L = 700 nm

Emin = (6.626 e-34*3e8)/(700 e -9)

Emin = 1.77 eV

so range of eenrgy is 1.77 , 3.105 eV

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