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

User Maraswrona
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1 Answer

3 votes

Answer:

The spectral half-width is 11.1 nm.

Step-by-step 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.

User Atul Kakrana
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