Answer to Question #88790 in Molecular Physics | Thermodynamics for Shivam Nishad

Question

Write Planck's formula for energy density of a
black body radiation. Show that Rayleigh-Jeans
law, Wien's law and Stefan's law are contained
in it

Answer 1

The Planck's formula for spectral energy density has the following form:

$$\rho_{\omega} = \frac{\hbar \omega^3}{\pi^2 c^3} \frac{1}{e^{\frac{\hbar \omega}{kT}} - 1}$$

1) Rayleigh-Jeans law can be obtained as a low-frequency limit of the Planck's formula:

$$\frac{\hbar \omega}{k T} << 1, \quad e^{\frac{\hbar \omega}{kT}} \approx 1 + \frac{\hbar \omega}{k T}$$

$$\Rightarrow \quad \rho_{\omega} \approx \frac{\hbar \omega^3}{\pi^2 c^3} \frac{1}{\frac{\hbar \omega}{kT}} = \frac{\omega^2}{\pi^2 c^3} kT$$

2) in order to derive Wien's law, one should obtain the Planck's formula in terms of a wavelength. This can be done by means of the following transformation:

$$\rho_{\omega} d\omega = \rho_{\lambda} d\lambda \, \Rightarrow \, \rho_{\lambda} = \rho_{\omega} \left|\frac{d\omega}{d\lambda} \right|,$$

where we take into account that increase in a frequency means decrease in a wavelength by putting the absolute value. Taking into account that

$$\omega = \frac{2 \pi c}{\lambda}, \quad d\omega = - \frac{2 \pi c}{\lambda^2} d\lambda,$$

we derive

$$\rho_{\lambda} = 16 \pi^2 c \hbar \frac{1}{\lambda^5 (e^{\frac{2\pi c\hbar}{kT \lambda}}-1)}$$

The maximum value of this function can be obtained from the condition

$$\frac{d \rho_{\lambda}}{d\lambda} = 0$$

Approximate solution of this equation leads to:

$$\frac{2 \pi c \hbar}{k T \lambda_{max}} = 4.965 \, \Rightarrow \, \lambda_{max} T = \frac{2 \pi c \hbar}{4.965 k} \equiv b$$

3) finally, Stefan-Boltzmann's law can be obtained by means of calculating the energy density function as

$$U = \int_0^\infty \rho_{\omega} d\omega = \frac{\hbar}{\pi^2 c^3} \int_0^\infty \frac{\omega^3 d\omega}{e^{\frac{\hbar \omega}{kT} }-1} =\frac{\pi^2 k^4}{15 c^3 \hbar^3} T^4 \equiv \alpha T^4.$$