# Deriving Planck's Constant from Wien's Displacement Law

So I'm reading an introductory book on Quantum Theory (David Park, 3rd ed.) and I am having trouble with the following question:

"According to Wien's displacement law, the wavelength $$\lambda_m$$ at which blackbody radiation at temperature T has its maximum intensity is given roughly by $$\lambda_mT \simeq$$ 3 mm K. Assuming that the quantum energy at this temperature is of the order of kT where k is Boltzmann's constant, estimate the value of Planck's constant."

My attempt at a solution is as follows:

$$\lambda_mT \simeq$$ 3 mm K $$,\quad$$ $$\lambda = \frac c \nu$$ $$\quad \rightarrow\quad$$ $$\frac {cT} \nu = 3\cdot10^{-3} mK$$ $$\cdot (\frac k c )\quad \rightarrow \quad$$ $$\frac {kT} \nu = 1.38\cdot10^{-34}Js$$

$$(\frac 1 2 mv^2)_{max} = kT_{max} = h\nu - e\phi$$

$$\frac {kT_{max}} \nu + \frac {e\phi} \nu =h$$

$$1.38\cdot10^{-34} + \frac {e\phi} \nu = h$$

At this point I get stuck, the orders of magnitude and units seem to be right but I'm not sure what to invoke/what is given in the question that can solve/eliminate this last term to get h, or if I'm even on the right track.

## 1 Answer

You do not need to convert the electromagnetic energy into the energy of a particle. You just posit that a photon with wavelength $$\lambda_{m}$$ has energy roughly $$kT$$; then you get that $$h\approx ckT/\lambda_{m}\approx\times10^{-34}$$ kg$$\cdot$$m$$^{2}\cdot$$s$$^{-1}$$, which is the correct order of magnitude. (Note that you only have one significant figure accuracy in the quantities you are given.)

If you want better than order of magnitude precision, the numerical calculation of the precise constant of proportionality between $$kT$$ and $$hc/\lambda$$ is outlined on the Wikipedia page for Wein's Law.