Use the Planck Law (1.10) to derive the Wien law for blackbody radiation. Hint: You will obtain

Use the Planck Law (1.10) to derive the Wien law for...

Key Concepts

Planck's Law

Planck's Law is a fundamental principle that describes the spectral distribution of electromagnetic radiation emitted by a blackbody in thermal equilibrium. It expresses the radiance as a function of wavelength (or frequency) and temperature, incorporating quantum concepts to accurately predict the observed spectrum at all wavelengths.

Blackbody Radiation

Blackbody Radiation refers to the idealized emission of electromagnetic radiation by an object that absorbs all incident radiation. The shape of its spectrum is uniquely determined by its temperature, making it a critical model for understanding thermal radiation in physics and astronomy.

Wien's Displacement Law

Wien's Displacement Law establishes that the wavelength at which a blackbody's emission is maximized is inversely proportional to its temperature. This law provides a simple relation to determine the temperature of an object by measuring the peak wavelength of its emitted spectrum.

Maximization of the Spectral Radiance

Deriving Wien's Law from Planck's Law involves differentiating the spectral radiance function with respect to wavelength or frequency and setting the derivative equal to zero. This maximization process is used to determine the point where the radiation intensity peaks, leading to the formulation of Wien’s Law.

Numerical Methods

The derivation of Wien's Law from Planck's Law results in a transcendental equation that cannot be solved analytically. Instead, numerical methods are employed to approximate the solution of this equation, illustrating the importance of computational techniques in solving complex equations in physics.

Transcript

00:01 The weem displacement law was discovered through empirical manner, where it was discovered that lambda, which is the wavelength maximum of the emission of a black body radiator, multiplied by the temperature of that radiator, is equal to this constant.

00:21 We're going to take planck's theoretical description for black body radiation, it's the distribution law, and differentiate it with respect to lambda to show that we will end up with this empirical law.

00:39 This is going to require a few math steps that i've written ahead of time.

00:48 So as we move down here, we can see how we're going to start.

00:53 First thing we need to write is what is planck's distribution law? this distribution function of the wavelengths multiply which, sorry, as a function of temperature.

01:09 So we have the distribution as a function of temperature multiplied by the range.

01:14 This portion here is how a distribution function is written.

01:19 It depends on 8 pi hc divided by lambda to the fifth, multiplied by d lambda, divided by this exponential function that includes hc over lambda kb t minus 1.

01:35 We were asked to, or hinted to use a substitution, so we'll do that.

01:43 So i rewrote this equation the right side down here, but we're going to take the derivative with respect to lambda.

01:51 So this portion is going to become 1.

01:53 D -lamda divided by d -lamda is 1.

01:57 So we're left with the rest of the function the same.

02:02 And i've substituted in the x value as hinted.

02:06 And the same equation is here, but now we have e to the x minus 1.

02:13 So we're going to differentiate this.

02:16 This is going to take some math.

02:18 There's two problems.

02:20 We have two portions of this equation that depend on lambda.

02:27 So this lambda, we're taking the derivative with respect to lambda.

02:30 We have lambda to the negative 5 here.

02:33 But remember our x depends on lambda.

02:36 There's a lambda in the denominator.

02:38 So we need to use, the product rule and the chain rule.

02:44 First we're going to tackle this side.

02:46 We're going to need the chain rule for this side because we have an exponential with the function.

02:56 So here is the chain rule.

03:02 The chain rule says that if you have a function that depends on an exponential of x, what we do is we first take the derivative of the entire function, so that's this part, derivative of the whole function, treat it as your variable.

03:26 So this whole function, e to the a x minus 1, the derivative of that part, as an example, is negative 1 divided by e to the a x minus 1 squared.

03:39 Then we take the derivative of the inside, so the e to the ax minus 1, that derivative is a to the eax, and we have chained through and multiplied them.

03:53 So we have set this part equal to our y and taken the derivative.

04:02 So now looking at the equation, this is our same equation, we're taking the derivative of this.

04:10 Before we apply the chain rule, we need to start with the product rule.

04:16 So because both functions here depend on lambda, we need to go through the product rule, which states if you have a derivative of a function where two parts are multiplied together and they both depend on this variable, you leave you, the first portion constant.

04:36 So in our case, this part is a and this part is b.

04:47 So we can take out the constant here, 8 pi hc here, no dependence on lambda, and we're going to say a y, so this is the first part, 1 over lambda to the 5th because the 8 pi hc is removed, and we have to take the derivative of the second part, the b.

05:12 And multiplied by the derivative of part 2.

05:16 Now we're going to add that.

05:18 Now the second part is constant, and we're going to take the derivative of part 1.

05:24 The derivative, remember we've pulled out this constant, so the derivative is the derivative of lambda to the negative 5.

05:34 So we've set it up, and now, when we take the derivative of this x1 divided by e to the x minus 1, we're going to insert the chain rule here.

05:50 So let's write this out.

05:52 So we have our constant in the front, 1 over lambda to the fifth.

05:57 Now i've plugged in the derivative based on the chain rule.

06:02 Remember we took the derivative of e to the x minus 1, negative 1...