We're learning about emission and absorbtion spectroscopy, and how an emission spectrum is produced by passing a high voltage through a gas, and an absorbtion spectrum is produced by shining white light on a gas. Since both techniques seem to be inputting energy into the sample, why don't both produce an emission spectrum (as the electrons fall back down to their 'natural' energy state)? Also, why do the electrons even fall back to their original energy state - since a constant stream of energy is being delivered, why don't they stay at an elevated state until that stream of energy stops?

  • 1
    $\begingroup$ 1. After absorption there are quite many pathways by which the molecule goes back to the ground state; not all of them involve radiative decay i.e. decay with emission of a photon. 2. Absorption of energy isn't classical - it's quantum. So it's not the same as a ping pong ball with a hairdryer underneath. If you excite a molecule with light $\ce{M + $h\nu$ -> M^*}$, shining more photons with energy $h\nu$ on the excited state doesn't "prop" the excited state up. It could decay back down to the ground state and get re-excited, but it will decay, purely because it likes having lower energy, $\endgroup$ – orthocresol Nov 1 '16 at 0:31
  • $\begingroup$ Thanks for the clarification and the help! I didn't realize that I was thinking of electrons in the wrong way the whole time. $\endgroup$ – Andi Gu Nov 1 '16 at 0:41
  • 1
    $\begingroup$ To emphasize the point that @orthocresol made: suppose you have a transition between states $A$ and $B$ with energy difference $E$. If introduce photos with energy $E$, you will promote both the transition $A\rightarrow B$ and $B\rightarrow A$. This process is governed by the Einstein coefficients for absorption and stimulated emission. $\endgroup$ – Zhe Nov 1 '16 at 0:50

Your Answer

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

Browse other questions tagged or ask your own question.