The radiation emitted by a blackbody at temperature T has a...

The radiation emitted by a blackbody at temperature $T$ has a frequency distribution given by the Planck spectrum: $$ \epsilon_{T}(f)=\frac{2 \pi h}{c^{2}}\left(\frac{f^{3}}{e^{h f / k_{\mathrm{B}} T}-1}\right) $$ where $\epsilon_{T}(f)$ is the energy density of the radiation per unit increment of frequency, $v$ (for example, in watts per square meter per hertz), $h=6.626 \cdot 10^{-34} \mathrm{~J} \mathrm{~s}$ is Planck's constant, $k_{\mathrm{B}}=1.38 \cdot 10^{-23} \mathrm{~m}^{2} \mathrm{~kg} \mathrm{~s}^{-2} \mathrm{~K}^{-1}$ is the Boltzmann constant, and $c$ is the speed of light in vacuum. (We'll derive this distribution in Chapter 36 as a consequence of the quantum hypothesis of light, but here it can reveal something about radiation. Remarkably, the most accurately and precisely measured example of this energy distribution in nature is the cosmic microwave background radiation.) This distribution goes to zero in the limits $f \rightarrow 0$ and $f \rightarrow \infty$ with a single peak in between those limits. As the temperature is increased, the energy density at each frequency value increases, and the peak shifts to a higher frequency value. a) Find the frequency corresponding to the peak of the Planck spectrum, as a function of temperature. b) Evaluate the peak frequency at temperature $T=6.00 \cdot 10^{3} \mathrm{~K}$, approximately the temperature of the photosphere (surface) of the Sun. c) Evaluate the peak frequency at temperature $T=2.735 \mathrm{~K}$, the temperature of the cosmic background microwave radiation. d) Evaluate the peak frequency at temperature $T=300 . \mathrm{K}$, which is approximately the surface temperature of Earth.

The radiation emitted by a blackbody at temperature $T$ has a frequency distribution given by the Planck spectrum:
$$
\epsilon_{T}(f)=\frac{2 \pi h}{c^{2}}\left(\frac{f^{3}}{e^{h f / k_{\mathrm{B}} T}-1}\right)
$$
where $\epsilon_{T}(f)$ is the energy density of the radiation per unit increment of frequency, $v$ (for example, in watts per square meter per hertz), $h=6.626 \cdot 10^{-34} \mathrm{~J} \mathrm{~s}$ is Planck's constant, $k_{\mathrm{B}}=1.38 \cdot 10^{-23} \mathrm{~m}^{2} \mathrm{~kg} \mathrm{~s}^{-2} \mathrm{~K}^{-1}$ is the Boltzmann constant,
and $c$ is the speed of light in vacuum. (We'll derive this distribution in Chapter 36 as a consequence of the quantum hypothesis of light, but here it can reveal something about radiation. Remarkably, the most accurately and precisely measured example of this energy distribution in nature is the cosmic microwave background radiation.) This distribution goes to zero in the limits $f \rightarrow 0$ and $f \rightarrow \infty$ with a single peak in between those limits. As the temperature is increased, the energy density at each frequency value increases, and the peak shifts to a higher frequency value.
a) Find the frequency corresponding to the peak of the Planck spectrum, as a function of temperature.
b) Evaluate the peak frequency at temperature $T=6.00 \cdot 10^{3} \mathrm{~K}$, approximately the temperature of the photosphere (surface) of the Sun.
c) Evaluate the peak frequency at temperature $T=2.735 \mathrm{~K}$, the temperature of the cosmic background microwave radiation.
d) Evaluate the peak frequency at temperature $T=300 . \mathrm{K}$, which is approximately the surface temperature of Earth.

Key Concepts

Wien's Displacement Law

Wien's Displacement Law is the principle that relates the temperature of a blackbody to the frequency (or wavelength) at which the emission spectrum peaks. It shows that as the temperature increases, the peak of the spectral distribution shifts to a higher frequency (or shorter wavelength), providing a straightforward method for estimating the temperature of emitting bodies based on their radiation spectrum.

Quantum Hypothesis of Light

The quantum hypothesis of light is the concept that light is composed of discrete packets of energy called photons. This hypothesis was a key development in early quantum theory and is critical to understanding phenomena such as blackbody radiation, as it explains why energy can only be emitted or absorbed in specific quantities, which is central to the derivation of Planck's Law.

Blackbody Radiation

Blackbody radiation refers to the spectrum of electromagnetic radiation emitted by an idealized physical body that absorbs all incident light and re-emits energy solely based on its temperature. This concept is central to thermodynamics and quantum physics because it provides a model for understanding how objects emit radiation across different wavelengths and frequencies.

Planck's Law

Planck's Law is the fundamental equation that describes the spectral energy density emitted by a blackbody as a function of frequency (or wavelength) and temperature. This law introduced the idea of quantized energy levels and paved the way for quantum mechanics by accurately modeling how energy is distributed among different frequencies.

Transcript

00:01 Hi there.

00:01 So for this problem, we are told that the radiation emitted by a blood body at temperature t has a frequency distribution given by the planet spectrum.

00:12 That is the equation shown in here.

00:15 Where epsilon t is the energy density of the radiation per unit increment of frequency.

00:26 So for example, we are given the value in here for plan.

00:32 Comes constant and for boltzmann constant.

00:41 We are also told that the distribution goes to zero in the limits, when the frequency goes to zero, and when the frequency goes to infinity, with a single peak in between those limits.

00:59 As the temperature is increased, the energy density at each frequency value increases, and the peak shift to a higher frequency value.

01:08 So for part a of this problem, we need to find the frequency corresponding to the peak of the plank spectrum as a function of temperature.

01:21 So in order to do that, we need to, the frequency of the peak of the plank distribution is determined by solving the following equation.

01:36 That is the derivative of the function that we are given with respect to the frequency.

01:42 And then we set this equal to zero because we want to find a maximum.

01:48 With that set, the derivative is going to take the following form.

01:55 So as you can see in here, we have two functions that depend on the frequency.

02:08 This function, which is the frequency elevated to the three, and this function which is the exponential of the frequency.

02:18 With other terms.

02:21 But when we are derivating such a function, we need first to derive one of the functions leaving the other constant and then we do the opposite.

02:39 We just divide the other function and leave the other function constant.

02:44 So in this case, the result that we are going to obtain is.

02:50 So we obtain this expression right here.

02:56 And remember that we need to set this equal to zero.

02:59 As you can see, here corresponds to the derivative of the frequency to the three, and this corresponds to the derivative of the turn of the exponential.

03:11 So now, solving four and this equal to zero, what we are going to do is to rearrange some things in here to take taking out some turns that are common.

03:27 As you can see, we already take out the constants and because we pass this to multiply the other side, we just, we are just going to have this one whole turn in here.

03:46 What we can do there is to take out the turn of the turn of, this whole turn to the minus 2.

03:56 So if we do that, we will obtain the following.

03:59 We will obtain the frequency to the 2 times the term of the exponential minus 1, that elevated to the minus 2.

04:14 So if we take out that, we are left with 3 to the exponential minus 1, minus, minus, frequency, balsman constant, the temperator, and this times the exponential of this right here.

04:43 And all of this equal to zero.

04:47 So this leads to the following equation.

04:51 Here we are going to make some simplifications.

04:54 We are going to make that x is equal to the product between plants constant, the frequency, divided by boltzmann constant times the temperature.

05:06 So in that way, the expression that we obtain becomes the following, becomes three times the exponential of s minus 3 minus x times the exponential of x.

05:24 If i'm correct, and this is equal to 0.

05:29 So, simplifying this expression yells the following.

05:33 3 minus x.

05:35 Is equal to 3 times the exponential of minus adds...

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