Practice Problem 28.1 Let us explore the relationship between the energy and wav

Question

Practice Problem 28.1 Let us explore the relationship between the energy and wavelength of a photon. Silicon films become better electrical conductors when they are illuminated by photons with energies of 1.14 eV or greater. This behavior is called photoconductivity. What is the corresponding photon wavelength? In what portion of the electromagnetic spectrum does it lie? SOLUTION SET UP The energy E of a photon is related to its frequency f by E=hf. The wavelength is related to the frequency by the general wave relationship =c/f. Combining these relationships and solving for , we get =hc/E. SOLVE It’s easiest to perform our calculations using units of (eVs) for h; that is, h=4.136×1015eVs. Using the preceding expression for , we get ==hcE=(4.136×1015eVs)(3.00×108m/s)1.14eV1.09×106m=1090nm The wavelengths of the visible spectrum are about 400 to 700 nm, so the wavelength we have found is in the near-infrared 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 causes photoconduction for silicon—in this case, about 1090 nm. Thus, for silicon, all light with a wavelength shorter than 1090 nm, including all light in the visible spectrum, contributes to photoconductivity. Part A - Practice Problem: If we need a material that is photoconductive for any wavelength in the visible spectrum (i.e., 400 to 700 nm), to what range of photon energies must it respond? Express your answers in electron-volts to three significant figures, separated by a comma.

Explanation / Answer

wavelength and Eneergy are related by E = hc/L

so for wavelength L = 400 nm

E1 = (6.626 e-34*3e8)/(400 e -9)

E1 = 3.105 e V

for L = 700 nm

E2 = (6.626 e-34*3e8)/(700 e -9)

E2 = 1.77 eV

so range of eenrgy is 1.77 eV , 3.105 eV

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