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Stefan Boltzmann’s laws’ statement proves the direct relation between the net energy emitted or radiated per unit surface area of the black body to the fourth power of the thermodynamic temperature of the black body.

As per the Stefan-Boltzmann law, the amount of radiation (u) radiated from a black body per unit time from an area (A), at an absolute temperature (T) is directly associated with the fourth power of the temperature.

Mathematically, we can the Boltzmann Equation as:

u = sAT4 . . . . . . (1)

Here, s = Stefan’s constant = 5.67 × 10-8 W/m2 k4

If a body is not a black body, it will absorb the energy. Therefore, the body emits less radiation.

Mathematically, the expression for such a body is:

u = e σ AT4 . . . . . .. (2)

Here, e = emissivity (it lies between 0 to 1 and is equal to absorptive power)

If the temperature all around is T0, the total radiated energy per unit time will be:

Δu = u – uo = eσA [T4 – T04] -------(3)

This law helps to relate the black body’s temperature with its net emitted power per unit area.

Expressing the statement mathematically, we get:

ε = σT4

Boltzmann law depicts the amount of power emitted from a blackbody with respect to the temperature.

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Boltzmann constant is a physical constant used in thermodynamics. It is the constant that relates the average kinetic energy of the gas with its temperature. The unit of temperature is represented by k.

J/K or m2Kgs-2K-1 is the unit of measurement of the Boltzmann constant. The Boltzmann constant is mostly used in Planck’s law of black body radiation, and Boltzmann’s entropy formula.

In this article, we will learn about Boltzmann’s constant, the value of the Boltzmann constant in the SI unit, the Boltzmann equation, and Stefan- Boltzmann law of radiation.

It is a physical constant that is represented by sigma (σ). In the Stefan Boltzmann law, it is a constant.

Value of Boltzmann Constant

We can get the Boltzmann constant by dividing the gas constant (R) by Avogadro’s number (NA).

So, the value of Boltzmann constant (kB) = 1.3806452 × 10-23J/K

Value of Boltzmann Constant in SI Unit

In eV, the value of Boltzmann constant is: 8.6173303 × 10-5 eV/K

There are many units to express the value of Boltzmann’s constant. The table stated below contains the value of k along with various units:

We can obtain the total power radiated per unit area, overall wavelengths of a black body by integrating Plank’s radiation formula.

Now, as a function of wavelength, the radiated power per unit area is:

\[\frac{dP}{d\lambda}\]\[\frac{1}{A}\]=\[\frac{2\pi hc^{2}}{\lambda^{5}(\frac{hc}{e\lambda kT^{-1}})}\]

Here, P = radiated power

λ = wavelength of the emitted radiation

A = surface area of a blackbody

c = velocity of light

h = Planck’s constant

T = temperature

k = Boltzmann’s constant

After simplifying the above eqn, we get:

\[\frac{d(\frac{P}{A})}{d\lambda }\] = \[\frac{2\pi hc^{2}}{\lambda^{5}(\frac{hc}{e\lambda kT^{-1}})}\]

Let’s integrate both sides w.r.t. λ to get the result as:

\[\int_{0}^{\infty }\] \[\frac{d(\frac{P}{A})}{d\lambda }\] = \[\int_{0}^{\infty}\] [\[\frac{2\pi hc^{2}}{\lambda^{5}(\frac{hc}{e\lambda kT^{-1}})}\]]d\[\lambda\]

After separating the constants, the integrated power is:

\[\frac{P}{A}\] = 2\[\pi\]hc\[^{2}\] \[\int_{0}^{\infty}\] [\[\frac{d\lambda}{\lambda^{5}(\frac{hc}{e\lambda kT^{-1}})}\]] - (1)

Analytically, we can solve this by substituting:

x = \[\frac{hc}{\lambda kT}\]

Therefore,

dx = - \[\frac{hc}{\lambda^{2}kT}\]d\[\lambda\]

⇒h = \[\frac{x\lambda kT }{c}\]

⇒c = \[\frac{x\lambda kT }{h}\]

⇒d\[\lambda\] = - \[\frac{\lambda^{2}kT}{hc}\]dx

Now substitute the value d\[\lambda\] of in equation 1, we get:

⇒\[\frac{P}{A}\] = 2\[\pi\](\[\frac{x\lambda kT }{c}\])(\[\frac{x\lambda kT }{h}\])\[^{2}\] \[\int_{0}^{\infty}\] [\[\frac{(-\frac{\lambda^{2}kT}{hc})dx}{e^{x}-1}\]]

= 2\[\pi\] (\[\frac{x^{3}\lambda^{5}k^{4}T^{4}}{h^{3}c^{2}\lambda^{5}}\]) \[\int_{0}^{\infty}\] [\[\frac{dx}{e^{x}-1}\]]

= \[\frac{2\pi(kT)^{4}}{h^{3}c^{2}}\] \[\int_{0}^{\infty}\][\[\frac{x^{3}}{e^{x}-1}\]]dx…[2]

The above equations are related to the standard form of integral as:

\[\int_{0}^{\infty}\][\[\frac{x^{3}}{e^{x}-1}\]]dx = \[\frac{\pi^{4} }{15}\]

Now replacing the above answer in equation 2, we get:

⇒\[\frac{P}{A}\] = \[\frac{2\pi(kT)^{4}}{h^{3}c^{2}}\] \[\frac{\pi^{4} }{15}\]

⇒\[\frac{P}{A}\] = (\[\frac{2k^{4}\pi^{5}}{15h^{3}c^{2}}\])T\[^{4}\]

After simplifying it further, we get:

⇒ P/A = σ T4

Finally, the formula for Stephen Boltzmann law is:

⇒ ε = σT4

Here, ε = P/A

σ = (\[\frac{2k^{4}\pi^{5}}{15h^{3}c^{2}}\]) = (5.670 х 10\[^{8}\]\[\frac{watts}{m^{2}K^{4}}\])

The Boltzmann equation is applied in a number of ways, these are:

Conservation equations– To derive the fluid dynamic conservation laws for energy, momentum, charge, and mass.

Quantum theory & violation of particle number conservation- Applications in physical cosmology that includes the production of dark matter, in Big Bang nucleosynthesis, the formation of light elements, etc.

In general relativity and astronomy.

FAQ (Frequently Asked Questions)

Q1. About 16 Jm⁻²s⁻¹ Amount of Energy is Emitted from a Hot Black Body at a Wavelength of 20,000 Å. The Radiation Emits the Most Intense Radiation. After some Time, the Temperature Rises. It Helps to Emit the Radiation with a Wavelength of 10,000 Å. Find What will be the Total Energy Radiation from the Hot Black Body in Jm⁻²s⁻¹?

Ans: According to Stefan Boltzmann law statement:

e = sT⁴

After expanding the formula, we get:

e₁/e₂ = (T₁/T₂)⁴

⇒ e₂ = (T₂/T₁)⁴. e₁

= (20000/10000)⁴ * 16

= (2)⁴. 16

= 16 * 16

= 256 Jm⁻²s⁻¹

Q2. Emissivity is a What Type of Property of a Material?

Ans: Emissivity is a metal’s optical property. It defines the amount of radiated (emitted) light from the material in accordance with an amount that radiates the black body at the same temperature.

Q3. Write a Short Note on ‘Blackbody’.

Ans: A blackbody is a kind of idealized physical body that consumes all incident electromagnetic radiation despite its angle of incidence or frequency.

Q4. Give a List of Factors that Affect Emissivity.

Ans: The following points show the factors that affect the emissivity:

Material: Metals have a higher emissivity than non-metals. So, quality is important.

Surface: More emissivity can be obtained from a polished and shiny surface.

Wavelength: Emissivity varies with the change in wavelength.

Geometry: Multiple reflections on the surface help to initiate a higher emissivity.

Temperature: Emissivity increases with the increase in temperature.