Calculate the energy (in eV) associated with a photon of green light of waveleng

Question

Calculate the energy (in eV) associated with a photon of green light of wavelength 0.5 mu m. Also, calculate the frequency (in Hz) of that light. Briefly explain why metals are opaque to electromagnetic radiation having photon energies within the visible region (lambda = 400-700 nm) of the electromagnetic spectrum, yet are transparent to high-frequency X- and y-ray radiation. Zinc telluride has a band-gap of 2.26 eV. Over what range of wavelengths of visible light (i.e. 0.4-0.7 pm) is it transparent?

Explanation / Answer

Here

energy of photon E = h * c/(wavelength * e)

energy of photon = 6.626 *10^-34 * 3 *10^8/(0.5 *10^-6 * 1.602 *10^-19)

energy of photon = 2.48 eV

the energy of photon is 2.48 eV

frequency * wavelength = speed of light

frequency * 0.5 *10^-6 = 3 *10^8

frequency = 6 *10^14 Hz

the frequency of that light is 6 *10^14 Hz

b)

as the high freqnecy em waves have higher energy and small waveelength , hence , they pass easily through metals

c)

energy = 2.26 eV

energy of photon E = h * c/(wavelength * e)

2.26 = 6.626 *10^-34 * 3 *10^8/(wavelength * 1.602 *10^-19)

wavelength = 0.549 um

the range of transparent wavelength is 0.549 um - 0.7 um

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