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

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.

ill do it for 100, also this isnt 80 its 40

false br