We know that a wave which has greater frequency will have low wavelength and high energy. So, by decreasing the wavelength, the frequency and consequently energy (intensity) of that wave will increase or vice versa.

Now, i want to ask a question in black body radiation by looking at the following graphs

enter image description here

My question is that if by decreasing the wavelength of a wave, the energy of that wave increases then why do these graphs fall down after reaching a maximum value. Shouldn't the graph just move straight on?

  • $\begingroup$ You seem to be confusing wavelength with amplitude. The graph depicts intensity, that goes with the (square of the) amplitude. Additionally, a blackbody happen to have that dependency of intensity with wavelength, but that is another history. There are another different shapes, like that of synchrotron radiation or when there are emission lines. $\endgroup$ – Eduardo Guerras Valera Jan 30 '13 at 1:19

The graph tells you how much radiation (intensity) of each wavelength is produced depending on the temperature. At 3000K for example, you wouldn't expect to see much low wavelength radiation compared to a hotter temperature. At 6000K however, which is much hotter, you do expect to see more high-energy, low wavelength radiation. This is what the graph is showing. As temperature increases, the peak wavelength is decreasing. This makes sense, because photons of low wavelength have higher energy.

The graph is showing that at a certain temperature, there is a distribution of a various different photons of various different energies. The peak of the distribution tells you which of these photons occurs the most at the given temperature.

| cite | improve this answer | |
  • $\begingroup$ thats what i'm saying. if photons having low wavelength have higher energy then why is their an abrupt fall in the graph after maximum intensity? the wavelength still decreases. $\endgroup$ – Rafique Jan 30 '13 at 2:12
  • 2
    $\begingroup$ @MuhammadRafique It is not a graph of photon energy as a function of wavelength. It is closer to a graph of number of photons as a function of wavelength. At shorter wavelength, yes a given photon will have more energy, but past a certain point a blackbody will emit very few such high-energy photons. $\endgroup$ – user10851 Jan 30 '13 at 4:31
  • $\begingroup$ so, can we say that the no of photons contained in a radiated wave after that maximum radiation point is little and hence the energy of the wave is little? $\endgroup$ – Rafique Jan 30 '13 at 15:41

You are confusing two very different concepts (but you are certainly not the first) whose biggest connection is that they are both named after Max Planck.

First, there is the energy of a single photon: $$ E = h\nu = \frac{hc}{\lambda}. $$ If you plotted $E$ as a function of $\lambda$, you would indeed get the monotonic relation you seek. This is the Planck relation.

On the other hand, we have the spectrum of light coming from a glowing hot extended object, given either by $$ B_\nu = \frac{2h\nu^3}{c^2(\mathrm{e}^{h\nu/kT}-1)} $$ (for power per unit area per unit frequency) or $$ B_\lambda = \frac{2hc^2}{\lambda^5(\mathrm{e}^{hc/\lambda kT}-1)} $$ (for power per unit area per unit wavelength). This is Planck's Law, which is what you have plotted. These are the distributions of energies (per unit time per unit area) of the many photons emitted from any object that glows (meaning any object above $0~\mathrm{K}$). We call such things spectra. You can divide them by the energy per photon given in the first formula, and thus get the distribution of the number of photons as a function of either frequency or wavelength.

| cite | improve this answer | |
  • $\begingroup$ so, is this graph between the quantity of radiation vs wavelength? quantity of radiation is maximum at a certain wavelength but decreases after that point, am i right? $\endgroup$ – Rafique Jan 30 '13 at 11:39
  • 1
    $\begingroup$ youtube.com/watch?v=ErRhupNgFS8 . This video explains my question but i dont get it $\endgroup$ – Rafique Jan 30 '13 at 14:37
  • $\begingroup$ this video explains it all.. there is not sufficient energy to produce high intensity-high frequency radiation $\endgroup$ – KawaiKx Feb 6 '17 at 5:42

I think you are confusing several aspects here. First, the energy of a photon depends on its frequency (or wavelength), the overall energy of a wave depends then on its frequency and number of photons.

The black body radiation law just says how much energy gets emitted by a black body of a given temperature at a given frequency. For example, consider an ordinary light bulb, which is a black body in a rather good approximation. Its spectrum will look pretty much like the black body radiation in your figure. It will emit some red photons, some blue, some infrared and even some ultraviolet. But when you count all the photons of a given frequency and multiply this number by energy of one photon, you'll get the same value as is in the graph. So what makes the biggest difference here is the different number of photons at different frequencies.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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