Question

Calculus Problem for fun? :D

Answer 1

Any body radiates energy at various wavelengths. Figure 4.118 shows the intensity of the radiation of a black body at a temperature T= 3000 kelvins as a function of the wavelength. The intensity of the radiation is the highest in the infrared range, that is, at wavelengths longer than that of visible light (0.4-0.7 micrometers). Max Planck's radiation law, announced to the Berlin Physical Society on October 19, 1900, states that

\[r(\lambda)=\frac{ a }{ \lambda^5(e ^{b/\lambda}-1) }\]

Find constants a and b so that the formula fits the graph (Later in 1900 Planck showed from theory that
\[a=2\pi c^2h \]
&
\[b=\frac{ hc }{ Tk }\]
h=Planck's constant
c= speed of light
k=Boltzmann's constant

Answer 2

@sunnnystrong
so is that an information or question you are stuck in?

Answer 3

Just posting a question to see who answers for a medal hahah :P

Answer 4

So what is the question?

Answer 5

@3mar up there ^^

Answer 6

"Find constants a and b so that the formula fits the graph"???

Answer 7

Welpp this is just an idea
The derivative of that function must be 0 at (0.96,3.13) according to that graph
Plugging in those coordinates in the derivative of the function we get 1 equation in terms of a and b

And the limit of the function as lambda approaches 0 is 0 so using this we get one more equation in terms of a and b

Now we have two equations in terms of a and b we can find a and b from em

Answer 8

The above procedure is fine, though I'm not so sure about the limit part being relevant. You can just solve the system
\[\begin{cases}
\dfrac{a}{0.96^5(e^{b/0.96}-1)}=3.13&\text{given point}\\[1ex]
\dfrac{a(b e^{b/0.96}-5(e^{b/0.96}-1)(0.96)}{0.96^7(e^{b/0.96}-1)^2}=0&\text{derivative at given point}
\end{cases}\]for \(a,b\). You should get something like \(a\approx363.23\) and \(b\approx4.76651\).