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I saw this equation today when calculating energies of photons of different frequencies, and noticed that the change in energy is a product of plank's constant and frequency. $$\Delta e = h * \nu $$ but what is the change in energy with respect to? since a change would be having some initial state $ e_f - e_i$ e sub f being final, e sub i being initial. What are these two initial and final states?

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  • $\begingroup$ The delta is between the two photons. $\endgroup$
    – user121330
    Commented Sep 9, 2014 at 22:10
  • $\begingroup$ What two photons? I was calculating the energy of one photon of UV radiation and there was no mention of another photon. $\endgroup$
    – AlanZ2223
    Commented Sep 9, 2014 at 22:12
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    $\begingroup$ Very odd to use $\Lambda$ as a frequency. Does your instructor do this? Just curious. $\endgroup$
    – BMS
    Commented Sep 9, 2014 at 22:33
  • $\begingroup$ I apologize I confused the variables with wavelength. Thanks for pointing that out. $\endgroup$
    – AlanZ2223
    Commented Sep 9, 2014 at 23:03
  • $\begingroup$ @AlanZ2223 The 'photons of different frequencies' that you mention in the first sentence? $\endgroup$
    – user121330
    Commented Sep 10, 2014 at 18:26

2 Answers 2

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For simplicity let's imagine a hydrogen atom, which has one proton and one electron.

The energy difference $\Delta E$ you speak of is the difference in energy of the two initial and final states of the electron-proton system.

Initially the system is in some state $\psi_i$ with energy $E_i$. Afterward, it's in a different state $\psi_f$ with energy $E_f$. In the language of non-relativistic quantum mechanics, these initial and final energies are eigenvalues of the time independent Schrodinger equation.

If all of this sounds completely foreign, I recommend getting a book on introductory quantum mechanics.

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I was confused about this but it turns out that the context of this equation is important. The energy change from an electron in one energy state to another can be determined by using the formula $\Delta E = h * \nu$ which will give the energy difference of the two states or the energy of the emitted or absorbed photon. However if you are just determining the energy of a photon without regards to a change in a system you can just use the equation without $\Delta$ so then it becomes $E = h * \nu$ which of course is the energy change of the particular quantum jump. So either formula just a way of looking at the same quantum jump from two different perspectives.

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