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Chapter 34: Problem 72

Show that Wien's law (Equation \(34.2 \mathrm{a})\) follows from Planck's law (Equation 34.3 ). (Hint: Differentiate Planck's law with respect to wavelength.)

Short Answer

Expert verified
By differentiating Planck's law with respect to wavelength, we can find the specific wavelength that leads to maximum spectral brightness, labelled as \(\lambda_{max}\). Multiplying this by the temperature, we find that it equals Wien's constant, thus proving Wien's law.

Step by step solution

01

Revisiting the laws

Wien's Law can be presented as \( \lambda_{\mbox{max}} T = k_{w}\), where \(\lambda_{\mbox{max}}\) is the wavelength at which the most radiation is emitted, \(T\) is the absolute temperature, and \(k_{w}\) is Wien's constant. Planck's radiation law is given by equation: \[ I(\lambda, T) = \frac{2hc^2}{\lambda^5}\frac{1}{e^{\frac{hc}{\lambda k_b T}} - 1}\] where: - \(I(\lambda, T)\) is the amount of energy per unit surface per unit time per unit solid angle emitted in the wavelength interval,- \(T\) is the temperature, - \(h\) is Planck's constant,- \(c\) is the speed of light,- \(k_b\) is Boltzmann constant.
02

Differentiate Planck's radiation law respecting wavelength

Differentiation of Planck's law with respect to \(\lambda\) gives:\[ \frac{dI(\lambda, T)}{d\lambda} = 0\]We then solve the equation for \(\lambda\), this will give us the wavelength at which the spectral brightness is highest. That's what we usually denote by \(\lambda_{max}\).
03

Complete the proof

Once you find the value of \(\lambda_{\mbox{max}}\), multiply it by temperature \(T\). You'll find that \(\lambda_{\mbox{max}} T = k_{w}\) , which is the expression of Wien's Law. Hence, proving that Wien's law follows from Planck's law.

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