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.

nec chapter 5 contains

c. information on special occupancy areas

nec chapter 6 covers

a. special equipment

which of the following rules contain the words shall or shall not and must be adhered to?

b. mandatory rules

which of the following rules identify actions that are allowed but not required and typically cover options or alternative methods?

c. optional rules

which of the following serves as the introduction to the nec and provides an overview of the book?

a. nec article 90

organizations that test products for safety are

c. nrtls

the nec is published by the

a national fire protection association

the nec is divided into

b. 9 chapters

the nec provides informative annexes for

c. informational materials and examples

two terms from nec article 100 that trainees should become familiar with are

b. labeled and listed

explanation:

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86%

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