# 29.3 Photon energies and the electromagnetic spectrum  (Page 4/16)

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## Visible light

The range of photon energies for visible light    from red to violet is 1.63 to 3.26 eV, respectively (left for this chapter’s Problems and Exercises to verify). These energies are on the order of those between outer electron shells in atoms and molecules. This means that these photons can be absorbed by atoms and molecules. A single photon can actually stimulate the retina, for example, by altering a receptor molecule that then triggers a nerve impulse. Photons can be absorbed or emitted only by atoms and molecules that have precisely the correct quantized energy step to do so. For example, if a red photon of frequency $f$ encounters a molecule that has an energy step, $\Delta E,$ equal to $\text{hf},$ then the photon can be absorbed. Violet flowers absorb red and reflect violet; this implies there is no energy step between levels in the receptor molecule equal to the violet photon’s energy, but there is an energy step for the red.

There are some noticeable differences in the characteristics of light between the two ends of the visible spectrum that are due to photon energies. Red light has insufficient photon energy to expose most black-and-white film, and it is thus used to illuminate darkrooms where such film is developed. Since violet light has a higher photon energy, dyes that absorb violet tend to fade more quickly than those that do not. (See [link] .) Take a look at some faded color posters in a storefront some time, and you will notice that the blues and violets are the last to fade. This is because other dyes, such as red and green dyes, absorb blue and violet photons, the higher energies of which break up their weakly bound molecules. (Complex molecules such as those in dyes and DNA tend to be weakly bound.) Blue and violet dyes reflect those colors and, therefore, do not absorb these more energetic photons, thus suffering less molecular damage.

Transparent materials, such as some glasses, do not absorb any visible light, because there is no energy step in the atoms or molecules that could absorb the light. Since individual photons interact with individual atoms, it is nearly impossible to have two photons absorbed simultaneously to reach a large energy step. Because of its lower photon energy, visible light can sometimes pass through many kilometers of a substance, while higher frequencies like UV, x ray, and $\gamma$ rays are absorbed, because they have sufficient photon energy to ionize the material.

## How many photons per second does a typical light bulb produce?

Assuming that 10.0% of a 100-W light bulb’s energy output is in the visible range (typical for incandescent bulbs) with an average wavelength of 580 nm, calculate the number of visible photons emitted per second.

Strategy

Power is energy per unit time, and so if we can find the energy per photon, we can determine the number of photons per second. This will best be done in joules, since power is given in watts, which are joules per second.

Solution

The power in visible light production is 10.0% of 100 W, or 10.0 J/s. The energy of the average visible photon is found by substituting the given average wavelength into the formula

$E=\frac{\text{hc}}{\lambda }.$

This produces

$E=\frac{\left(6\text{.}\text{63}×{\text{10}}^{\text{–34}}\phantom{\rule{0.25em}{0ex}}\text{J}\cdot \text{s}\right)\left(3.00×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m/s}\right)}{\phantom{\rule{0.25em}{0ex}}\text{580}×{\text{10}}^{\text{–9}}\phantom{\rule{0.25em}{0ex}}\text{m}}=\text{3.43}×{\text{10}}^{\text{–19}}\phantom{\rule{0.25em}{0ex}}\text{J}.$

The number of visible photons per second is thus

$\text{photon/s}=\frac{\text{10.0 J/s}}{3\text{.}\text{43}×{\text{10}}^{\text{–19}}\phantom{\rule{0.25em}{0ex}}\text{J/photon}}=\text{2.92}×{\text{10}}^{\text{19}}\phantom{\rule{0.25em}{0ex}}\text{photon/s}.$

Discussion

This incredible number of photons per second is verification that individual photons are insignificant in ordinary human experience. It is also a verification of the correspondence principle—on the macroscopic scale, quantization becomes essentially continuous or classical. Finally, there are so many photons emitted by a 100-W lightbulb that it can be seen by the unaided eye many kilometers away.

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for the answer to complete, the units need specified why
That's just how the AP grades. Otherwise, you could be talking about m/s when the answer requires m/s^2. They need to know what you are referring to.
Kyle
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my name is Abu m.konnek I am a student of a electrical engineer and I want you to help me
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