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31 May, 11:21

Calculate the spectral half-width at room temperature of an infrared LED of peak wavelength 550 nm.

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  1. 31 May, 12:03
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    The spectral half-width is 11.1 nm.

    Explanation:

    we know that if f is the frequency and h is the planck constant, then the energy is given by:

    E = h*f

    but if c is the speed of light and λ is the wavelength then, f = c/λ.

    E = [h*c]/λ

    λ = [h*c]/E

    then the change in λ is given by:

    Δλ = [ (h*c) / (E^2) ]*ΔE

    = [ (h*c) / ((h*c/λ) ^2) ]*ΔE

    = [ (λ^2) / (h*c) ]*ΔE

    but we also know that:

    ΔE = 1.8*k*T, where k = 1.38*10^-23J/K and T = 20 + 273 = 293K

    then the half-width is given by:

    Δλ = [ (λ^2) / (h*c) ]*1.8*kT

    = [ ((550*10^-9) ^2) / ((6.63*10^-34) * (3*10^8)) ]*1.8 * (1.38*10^-23) * (293)

    = 1.11*10^-8 m

    ≈ 11.1 nm

    Therefore, the spectral half-width is 11.1 nm.
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