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.