## Applicability of simplified blackbody radiation law (Quantum Physics)

Consider the black body radiation law. Simplify this expression considering $λT$ is sufficiently small that is, $exp⁡(h c/λkT)>1$.
B. Show that the simplified approximation that you obtained applies when $λ$ is in the visible range and $T = 1000K$.
C. Suppose the surface are of a heating element on an electric stove is $20 in^2$, (and its temperature is 1200 K). What power is radiated between the wavelengths 690 nm and 700 nm. What power is radiated between the wavelengths 400 nm and 410 nm. What color does have the heating element? Why?

Planck law is (energy emitted per unit area and unit time or power emitted per unit area):
$u(λ,T)=(2πhc^2)/λ^5 *1/(exp⁡(h c/λKT)-1)$
If the exponential is greater than 1, that is $h c>λKT$ one can write
$exp⁡(x)=1+x+x^2/2!+⋯=1+x$
So that
$u(λ,T)=2πcKT/λ^4$
which is called the Rayleigh-Jeans law.

b) For T =1000 K the above condition is true when
$λ<h c/KT=(6.626*10^{-34}*3*10^8)/(1.38*10^{-23}*1000)=1.44*10^{-5} m=14.4 μm$
Since the visible spectrum is in the order of hundreds of nm (380-750 nm), the above condition is always satisfied.

c) Power radiated per surface area is
$P(690-700 nm)/S=\int_{690 nm}^{700 nm} 2πcKT/λ^4*dλ =-2πcKT/(3λ^3)|_{700 nm}^{690 nm}$

$P(690-700 nm)/S=\frac{2π*3*10^8*1200*1.38*10^{-23}}{3*10^{-27} }* (1/690^3 -1/700^3 )=1.338*10^6 W/m^2$

For $S=20 in^2=0.0129 m^2$   we have $P=1.338*10^6*0.0129=17260.0 W=17.26 kW$
In the interval 400-410 nm we have
$P(400-410 nm)/S=2.322*10^7 W/m^2$
which means $P=2.322*10^7*0.0129=299489.5 W=299.49 kW$

To determine the color of the heating element write first the Planck law as
$u(λ)=A/λ^5 *1/(exp⁡(B/λ)-1)$
And obtain its derivative
$d u/dλ=⋯=0$    (this is a very complicated expression)
From maximum of power radiated one can determine the color (wavelength λ) corresponding to a given temperature. They are tabulated on the web. For T =1200 K the corresponding color is dark red.

valentin68