Final answer:
Zinc sulfide (ZnS) is transparent to visible light within the energy range of approximately 0.83 eV to 1.65 eV, corresponding to wavelengths of approximately 380 nm to 760 nm.
Step-by-step explanation:
Visible light refers to the range of wavelengths that are visible to the human eye. The range of visible light wavelengths is typically considered to be between 380 nm (nanometers) to 760 nm. To determine which wavelengths of visible light are transparent for zinc sulfide (ZnS), we need to find the corresponding range of photon energies that fall within the band gap of 3.66 eV.
The energy of a photon can be calculated using the equation E = hc/λ, where E is the energy of the photon, h is Planck's constant (6.626 x 10^-34 J·s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength of light in meters.
To convert the wavelength range of visible light into meters, we can use the conversion factor 1 nm = 1 x 10^-9 m. So, the range of visible light wavelengths in meters would be 380 x 10^-9 m to 760 x 10^-9 m.
Plugging these values into the energy equation:
E1 = (6.626 x 10^-34 J·s)(3.00 x 10^8 m/s)/(380 x 10^-9 m) = 1.65 eV
E2 = (6.626 x 10^-34 J·s)(3.00 x 10^8 m/s)/(760 x 10^-9 m) = 0.83 eV
Therefore, zinc sulfide (ZnS) would be transparent to visible light within the energy range of approximately 0.83 eV to 1.65 eV, corresponding to wavelengths of approximately 380 nm to 760 nm.