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Aneli [31]
3 years ago
8

When considering light moving through a diffraction grating, you should treat the light as what?

Chemistry
2 answers:
scZoUnD [109]3 years ago
8 0

Answer:

Wave

Explanation:

The dual nature of light is manifested through its behavior as a particle and as a wave depending on the experimental set-up.

Here, a diffraction grating is an optical component which splits falling on it, into characteristic wavelengths. The grating equation is given as:

n\lambda =2dsin\theta

where n = diffraction order 1,2,3...

λ=wavelength of light

d=distance between the grooves (lines) on the grating

θ=dispersion angle

since the wavelength of light (λ, in nanometers) is comparable to the length of the medium with which the light interacts which in this case is the value of 'd'; the light is treated as a wave.

Situations where the wavelength of light is not comparable with the length scale of the interaction medium, light is treated as a particle.

liraira [26]3 years ago
5 0

Answer:

A particle

Explanation:

Modern quantum theory holds that light has both wave-like and particle-like properties. When the length scales involved are large compared to the wavelengths of light (ex., forming images with thin lenses), the

particle nature of light dominates.

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Answer:

\Delta\text{H}_1+2\Delta\text{H}_2-\Delta\text{H}_3

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Hess's Law of Constant Heat Summation states that if a chemical equation can be written as the sum of several other chemical equations, the enthalpy change of the first chemical equation is equal to the sum of the enthalpy changes of the other chemical equations. Thus, the reaction that involves the conversion of reactant A to B, for example, has the same enthalpy change even if you convert A to C, before converting it to B. Regardless of how many steps it takes for the reactant to be converted to the product, the enthalpy change of the overall reaction is constant.

With Hess's Law in mind, let's see how A can be converted to 2C +E.

\bf{\text{A} \rightarrow 2\text{B}}                  (Δ\text{H}_1)  -----(1)

Since we have 2B, multiply the whole of II. by 2:

\bf{2\text{B} \rightarrow 2\text{C} +2\text{D}}       (2Δ\text{H}_2) -----(2)

This step converts all the B intermediates to 2C +2D. This means that the overall reaction at this stage is \text{A} \rightarrow 2\text{C} +2\text{D}.

Reversing III. gives us a negative enthalpy change as such:

\bf{2\text{D} \rightarrow \text{E}}                  (-Δ\text{H}_3) -----(3)

This step converts all the D intermediates formed from step (2) to E. This results in the overall equation of \text{A} \rightarrow 2\text{C} +\text{E}, which is also the equation of interest.

Adding all three together:

\text{A} \rightarrow 2\text{C}+\text{E}            (\bf{\Delta\text{H}_1+2\Delta\text{H}_2-\Delta\text{H}_3 })

Thus, the first option is the correct answer.

Supplementary:

To learn more about Hess's Law, do check out: brainly.com/question/26491956

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