(III) Planck's radiation law is given by: I(λ, T)=(2 πh c^2 λ^-5)/(e^h c / λk T-1) where I(λ, T)

(III) Planck's radiation law is given by: I(λ, T)=(2 πh c^2...

(III) Planck's radiation law is given by: $$ I(\lambda, T)=\frac{2 \pi h c^{2} \lambda^{-5}}{e^{h c / \lambda k T}-1} $$ where $I(\lambda, T)$ is the rate energy is radiated per unit surface area per unit wavelength interval at wavelength $\lambda$ and Kelvin temperature $T .(a)$ Show that Wien's displacement law follows from this relationship. (b) Determine the value of $h$ from the experimental value of $\lambda_{\mathrm{P}} T$ given in the text. [You may want to use graphing techniques.] (c) Derive the $T^{4}$ dependence of the rate at which energy is radiated (as in the Stefan-Boltzmann law , by integrating Planck's formul: over all wavelengths; that is, show that $$ \int I(\lambda, T) d \lambda \propto T^{4} $$

(III) Planck's radiation law is given by:
$$
I(\lambda, T)=\frac{2 \pi h c^{2} \lambda^{-5}}{e^{h c / \lambda k T}-1}
$$
where $I(\lambda, T)$ is the rate energy is radiated per unit surface
area per unit wavelength interval at wavelength $\lambda$ and
Kelvin temperature $T .(a)$ Show that Wien's displacement
law follows from this relationship. (b) Determine the value of $h$ from the experimental value of $\lambda_{\mathrm{P}} T$ given in the text.
[You may want to use graphing techniques.] (c) Derive the
$T^{4}$ dependence of the rate at which energy is radiated (as in the Stefan-Boltzmann law , by integrating Planck's formul:
over all wavelengths; that is, show that
$$
\int I(\lambda, T) d \lambda \propto T^{4}
$$

Key Concepts

Planck's Radiation Law

This law gives the distribution of electromagnetic radiation emitted by a blackbody as a function of wavelength and temperature. It marks a fundamental development in quantum theory by explaining how energy is quantized, resolving the ultraviolet catastrophe of classical physics, and forming the basis for further derivations in thermodynamics and radiation physics.

Wien's Displacement Law

This law states that the wavelength at which the emission of a blackbody peaks is inversely proportional to its absolute temperature. It is derived by finding the maximum of Planck's radiation function and illustrates the shift in the spectrum towards shorter wavelengths as the temperature increases.

Stefan-Boltzmann Law

This law expresses that the total energy per unit area radiated by a blackbody across all wavelengths is proportional to the fourth power of its absolute temperature. It is obtained by integrating Planck's radiation law over all wavelengths and serves as a key relation in thermodynamics and astrophysics.

Integration Techniques in Radiation

Deriving the T^4 dependence involves integrating the spectral radiance over the wavelength domain. This process connects the microscopic description provided by Planck's law to the macroscopic Stefan-Boltzmann law, emphasizing the role of mathematical integration in linking detailed spectral features to overall energy output.

Transcript

00:01 Now flanks here it is given plants, radiation law, intensity at given bay length and temperature 2 pi hc square lambda to the power minus 5 upon exponential of h c by lambda k t minus 1.

00:43 In the first part we have to show that wayne's displacement law is followed b part calculate the value of h using given value of lambda p and t c part show that i lambda t d lambda is directly version 2 t to the power 4 let us start solving it according to ways displacement law lambda p into t to be constant we have to find lambda p for intensity to be maximum so di upon d lambda must be zero that is a partial derivative because it is depending on the two function one is temperature and other is wavelength that should be zero so it can be written as after differentiation this can be written as 2 5 h c square it is to be constant exponential of s c upon lambda k t minus 1 into minus 5 lambda to the power minus 6 minus lambda to the power minus 5 exponential of sc by lambda y k t into 1 minus sc upon lambda t lambda square k t lambda square whole to be divided by e s c by lambda k t minus 1 whole square that is to be 0 so on solving it you will get s c by lambda k t cross multiply and simplify you will get e to the power x 5 minus x is called to 5 here let x to be s c by lambda k t so this term should be constant so s c by lambda k t is constant hence lambda p into t should be constant now b part to find value of constant we have to solve e to the power x 5 minus x is called to 5 or 5 minus x is equal to 5 e to the power minus x so on solving it value of x you will get 4 .996 and by calculated value you will get lambda p k t is called to 4 .966 so h will be 4 .966 lambda p k t upon c it can be written as 4 .966 k is biv 2 .5 lambda p into k is bench displacement constant 2 .9 0 10 to the bar minus 3 k is bolzman constant 1 .38 10 to the power minus 23 upon 3 into 10 to the power 8.

07:41 So it is to be 6 .62 10 to the power minus 34 june into second...

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