Assuming that your surface temperature is 98.6^∘ F and that you are an ideal blackbody radiator

Assuming that your surface temperature is 98.6^∘ F and that...

Assuming that your surface temperature is $98.6^{\circ} \mathrm{F}$ and that you are an ideal blackbody radiator (you are close), find (a) the wavelength at which your spectral radiancy is maximum, (b) the power at which you emit thermal radiation in a wavelength range of $1.00 \mathrm{~nm}$ at that wavelength, from a surface area of $4.00 \mathrm{~cm}^{2}$, and (c) the corresponding rate at which you emit photons from that area. Using a wavelength of $500 \mathrm{~nm}$ (in the visible range), $(\mathrm{d})$ recalculate the power and (e) the rate of photon emission. (As you have noticed, you do not visibly glow in the dark.)

Key Concepts

Photon Energy and Emission Rate

Each photon carries energy proportional to its frequency (or inversely proportional to its wavelength) as given by Planck's relation, E = hc/?. This concept is used to convert the power radiated in a narrow wavelength band into the number of emitted photons by dividing the total emitted power by the energy per photon, a critical step in understanding the quantized nature of thermal radiation.

Stefan-Boltzmann Law

The Stefan-Boltzmann Law connects the total power radiated per unit surface area of a blackbody to the fourth power of its temperature. While in this question the focus is on a specific spectral range rather than the total emission, the underlying principle of energy emission based on temperature originates from this law.

Blackbody Radiation

This concept describes an idealized physical body that absorbs all incident electromagnetic radiation and re-emits it with a characteristic spectrum that depends only on its temperature. In many physical scenarios, including the example at hand, objects approximate blackbody behavior, allowing us to use well-established laws of thermal radiation to predict emitted energy and distribution over wavelengths.

Wien's Displacement Law

Wien's Displacement Law relates the temperature of a blackbody to the wavelength at which its emission is maximized. It states that the product of the peak wavelength and the temperature is a constant, meaning that hotter objects emit peak radiation at shorter wavelengths. This law is fundamental when determining the spectral peak for a body at a given temperature.

Planck's Law

Planck's Law provides a description of the spectral radiance of a blackbody as a function of wavelength for a given temperature. It quantifies how much power is emitted at each wavelength and is essential for calculating the radiated power in a specific, narrow spectral band, as required in the problem.

Transcript

00:01 For this problem on the topic of photons, we want to assume that the human surface temperature is 98 .6 degrees fahrenheit and assume that it is an ideal black body radiator.

00:10 We want to find a wavelength at which the spectral radiance is a maximum, the power at which thermal radiation is emitted in a wavelength range of one nanometer at this wavelength from a surface area from a surface area of four square centimeters and the corresponding rate at which photons are emitted from that area.

00:26 Then for a wavelength of 500 nanometers, we want to recalculate the power as well as the rate of emission of photons.

00:32 Now, if we use a temperature t of 98 .6 degrees fahrenheit, this is 37 degrees celsius, which is 310 kelvin.

00:44 We can use wien's law and find the wavelength that corresponds to the spectral radiancy maximum, and we'll call this lambda max, and this is equal to 2 ,8009.

00:58 98 micrometer kelvin over the temperature t, which is 2 ,898 micrometers kelvin divided by 310 kelvin.

01:16 This gives the maximum or the wavelength that corresponds to the spectral radiancy maximum to be 9 .35 micrometers.

01:28 Now with the wavelength 9 .35 micrometers and t, 10 kelvin the spectral radiancy s of lambda we know is equal to 2 pi c squared h over lambda to the power 5 multiplied by 1 over e to the power h c over lambda k t minus 1 and so if we put values into this equation this is 2 pi times c squared which is 2 0 .998 times 10 to the 8 meters per second, and we'll suppress the units, and that's squared, times the clanks constant 6 .626 times 10 to the minus 34 joules seconds, divided by the wavelength, 9 .35 times 10 to the minus 6 meters, all to the power 5, multiplied by e to the power of 6 ,000, 626 times 10 to the minus 34 joules seconds times 2 .998 times 10 to the 8 meters per second divided by 9 .35 times 10 to the minus 6 meters multiplied by 1 .38 times 10 to the minus 23 jules per kent times 10 to the minus 6 meters multiplied by 1 .38 times 10 to the minus 23 joules per calvin times the temperature of 310 calvin.

03:15 And so doing this calculation, we get the spectral radiancy to be 3 .68 times 10 to the power 7 watts per cubic meter.

03:29 Now for a small range of wavelength, the radiated power may be approximated as p, which is the spectral radiancy s multiplied by the area a.

03:42 Times delta lambda.

03:46 And so this is 3 .68 times 10 to the power 7 watts per cubic meter times a surface area of 4 times 10 to the minus 4 square meters times a wavelength range of 1 nanometer or 10 to the minus 9 meters.

04:09 So calculating we get this power to be 1 .475.

04:15 Times 10 to the minus 5 watts.

04:22 And so the energy carried by each photon, epsilon is equal to hf, which is hc over lambda, which is 6 .626, times 10 to the minus 34 joules seconds times 2 .998 times 10 to the power 8 meters per second, divided by the wavelength 9 .35 times 10 to the minus 6 meters...

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