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.

ok

explanation:

jupiter is the largest planet in our solar system, having a mass and radius that are respectively, 318 and 11.2 times that of earth. suppose that an object falls from rest near the surface of each planet and that the acceleration due to gravity remains constant during the fall. each object falls the same distance before striking the ground. determine the ratio of the time of fall on jupiter to that on earth.