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). (d) recalculate the power and (e) the rate of photon emission. (As you have noticed, you do not visibly glow in the dark.)

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).
(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 are to assume that the body's surface temperature is 98 .6 degrees fahrenheit and that you are an ideal black body radiator.

00:11 We're going to find a wavelength at which your spectral radiancy is a maximum.

00:15 The power at which you emit thermal radiation in a wavelength range of a nanometer at that wavelength from a surface area of four square centimeters.

00:25 The rate at which you emit photons from that area and using a wavelength of four.

00:30 500 nanometers, we want to recalculate the power, the rate of photon, and the rate of photon emission.

00:38 Now, if we have a temperature t of 98 .6 degrees fahrenheit, this is equivalent to 37 degrees celsius, which is 310 kelvin.

00:48 We can use wien's law and find the wavelength that corresponds to the spectral radium, radency maximum, and we'll call this lambda max, and this is 2 ,800 micrometer calvin divided by the temperature t which is 2 ,898 micrometer calvin divided by 310 calvin which is a wavelength of 9 .35 micrometers.

01:32 For part b if we use the wavelength lambda as 9 .35 micrometers and the temperature t of 3 .3 5 micrometers.

01:42 310 kelvin, the spectral radiancy, s as a function of lambda, is equal to 2 pi c squared h divided by lambda to the power 5 times 1 over e to the power hc over lambda kt minus 1.

02:09 Now, if we put our values into this equation, this is 2 pi times 2 .998 times 10 to the power 8 meters per second squared, which is c, times 6 .626 times 10 to the minus 34 joules seconds, all divided by the wavelength 9 .35 times 10 to the minus 6 .6 meters or to the power of.

02:46 Multiplied by e to the power 6 .626 times 10 to the minus 34 joules seconds times 2 .998 times 10 to the power 8 meters per second, all divided by 9 .35 times 10 to the minus 6 meters times the balsman's constant 1 .38.

03:21 Times 10 to the minus 23 joules per calvin multiplied by a temperature of 310 kelvin.

03:34 And so calculating, we get the spectral radiance as to be 3 .688 times 10 to the power 7 watts per cubic meter.

03:48 For small range of wavelength, the radiated power may be approximated as p, which is the spectral radiancy s, multiplied by the area a times delta lambda...

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