7
$\begingroup$

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.

Improve this question
$\endgroup$
4
  • 4
    $\begingroup$ 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 :-) $\endgroup$
    – DanielSank
    Commented Mar 9, 2016 at 19:32
  • 1
    $\begingroup$ @DanielSank by fitting the maxima of his distribution to the experimental value of Wien Law, if I recall... $\endgroup$
    – AccidentalFourierTransform
    Commented Mar 9, 2016 at 19:35
  • $\begingroup$ 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 :-) $\endgroup$
    – DanielSank
    Commented Mar 9, 2016 at 19:53
  • $\begingroup$ Suggestion to the question formulation (v2): Replace the word calculate in various places with the word estimate. $\endgroup$
    – Qmechanic
    Commented Mar 9, 2016 at 20:05

1 Answer 1

Reset to default
8
$\begingroup$

See How did Planck derive the black body radiation formula without using the Bose statistics? for general circumstances of Planck's discovery. Specifically, Planck was trying to reproduce the expression for entropy of an ideal oscillator involved in the original derivation of the black body radiation law. To do so, he had to assume that energy is split into "droplets" and count the number of ways it could be distributed. The idea was to take continuous limit at the end. But that did not reproduce the expression Planck needed.

As an "act of despair" ("no matter what the cost, I must bring about a positive result") he fixed the energy of his droplets in proportion to the frequency of light, and introduced the constant of proportionality as a "new constant of nature". To get the value he fit it to the expression he had for the oscillator entropy, which in turn was fit to the experimental radiation measurements of 1899, the ones that showed that Wien's law, which Planck derived earlier, was not followed at low frequencies.

Improve this answer
$\endgroup$
4
  • $\begingroup$ This sounds fascinating but for the more mathematically-inclined users like myself, almost incomprehensible. Could you possibly fill in some details? $\endgroup$
    – Mikhail Katz
    Commented Mar 10, 2016 at 9:50
  • 1
    $\begingroup$ @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. $\endgroup$
    – KCd
    Commented Mar 10, 2016 at 13:28
  • 1
    $\begingroup$ @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 $1/a^2$. $\endgroup$
    – KCd
    Commented Mar 10, 2016 at 13:32
  • 2
    $\begingroup$ 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 $E_n = nh\nu$). $\endgroup$
    – KCd
    Commented Mar 10, 2016 at 13:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.