The thermal radiation is emitted over a broad spectrum of wavelengths. For a blackbody, this radiation follows the Planck radiation law:

$$P(\lambda) = \frac{2 \pi h c^{2}}{ \lambda^{5} \left( e^{\frac{hc}{\lambda k T}} – 1 \right) }$$

**P _{λ} is power emitted per unit area per wavelength interval at wavelength λ. **

– It is also called spectral emission

– SI units: W m

^{-2}m

^{-1}(Corresponding equal to power emitted per unit area multiplied by per wavelength interval)

**Importance of this formula: **

– Wien displacement law came from taking the derivative of this equation and setting to zero to find the maxima.

– Stefan-Boltzmann law came from integrating this equation over all wavelengths.

*Thermal Radiation*

*As temperature T increases:*

– the thermal peak wavelength shifts proportional to T^{-1}: Wien displacement law

– the total radiated power increases proportional to T^{4}: Stefan-Boltzmann law

The relation between the peak wavelength and the radiant body temperature is the Wien displacement law:

$$\lambda_{peak} T = 2.898 \times 10^{-3} \, m K$$

### Net Radiation Gain or Loss

The amount of electromagnetic energy absorbed by an object depends on the temperature of its surroundings. If an object at a temperature T and is fully embedded in a surrounding which is at temperature T_{surr}, the rate of net energy gained (or lost) by the object due to emission and absorption of electromagnetic radiation is:

$$\frac{dQ}{dt} = \sigma A e (T^{4} – T_{surr}^{4} )$$

Hence, if T > T_{surr}, the object radiates more energy than it absorbs, and its temperature decreases until equilibrium with its surroundings is reached.

Sir,I want to know how we find the value of hc/(lambda)kT.

Kindly help me to know the deduction Of the above ..