# How did Planck Calculate the Planck Constant?

Having started to learn about quantum behavior, this formula came up:

$$E = hf$$

Where $E$ is energy, $h$ is the Planck constant and $f$ is the frequency.

My physics teacher suggested an experiment involving LEDs as a way of calculating $h$.

However given that the Planck constant was calculated long before the invention of LEDs, how was it calculated? I know from doing some research Planck used something to do with black body radiation, but I cannot seem to find out the specifics of his method.

• Planck figured out that energy was quantised when trying to explain black body radiation. At the time, the classical theory predicted that a body would radiate an infinite amount of power, which is obviously wrong. Planck found that by assuming the mode excitations were quantized he got a convergent answer. None of this actually answers the question of how the value of $h$ was measured, which is why this is a comment :-) – DanielSank Mar 9 '16 at 19:32
• @DanielSank by fitting the maxima of his distribution to the experimental value of Wien Law, if I recall... – AccidentalFourierTransform Mar 9 '16 at 19:35
• The problem is that we're not entirely sure. What @AccidentalFourierTransform said sounds reasonable (i.e. I think it would work) but I don't know what actually happened historically. Frankly, I'd find this out by reading a few Wikipedia articles etc., which you could do just as fast as me :-) – DanielSank Mar 9 '16 at 19:53
• Suggestion to the question formulation (v2): Replace the word calculate in various places with the word estimate. – Qmechanic Mar 9 '16 at 20:05

• @katz, sums are like integrals, but for a parameter $a > 0$ compare $\int_0^\infty e^{-ax}\,dx = 1/a$ to $\sum_{n\geq 0} e^{-an} = 1/(1-e^{-a}) = e^a/(e^a-1)$. As $a \rightarrow 0^+$ the series behaves like $1/a$, which is the integral, but as $a \rightarrow \infty$ the series tends to 1 while the integral tends to $0$. This is the kind of dichotomy that happens between classical and quantum formulas when you replace an integral with a sum that both depend on a common parameter. – KCd Mar 10 '16 at 13:28
• @katz, as another example consider $\int_0^\infty xe^{-ax}\,dx = 1/a^2$ and $\sum_{n \geq 0} ne^{-an} = e^{-a}/(1 - e^{-a})^2 = e^a/(e^a-1)^2$. As $a \rightarrow 0^+$ the series behaves like $1/a^2$, which is the integral, but as $a \rightarrow \infty$ the series behaves like $1/e^a$, which decays much faster than $1/a^2$. – KCd Mar 10 '16 at 13:32
• The connection of the sums and integrals I wrote to physics is that, for blackbody radiation, the integral occurs with $a = 1/kT$ (allowed energies are continuous over all positive numbers) and the sum occurs with $a = h\nu/kT$ (allowed energies only at values $E_n = nh\nu$). – KCd Mar 10 '16 at 13:43