May 25, 2020, 01:50:29 AM
Forum Rules: Read This Before Posting


Topic: Bohr's equation, photons, and simple algebra.  (Read 8245 times)

0 Members and 1 Guest are viewing this topic.

Offline xiexieniii

  • Very New Member
  • *
  • Posts: 1
  • Mole Snacks: +0/-0
Bohr's equation, photons, and simple algebra.
« on: January 31, 2009, 05:24:04 PM »
My question is regarding Bohr's equation, which, from what I understand (please correct me if I'm wrong), is used to calculate the energy gained or lost when an electron changes "states" (n=1, n=2, etc). Not exactly sure what a "state" is...although I know that a change in "state" from higher to lower results in the emission of a photon. Are "states" the same as orbitals?

Anyway, my main question is how this equation went from step 1 to step 3:

/\E= Efinal - Einitial

    = (-2.18 x 10-18 J)(1/nf2) - (-2.18 x 10-18 J)(1/ni2)

    = -2.18 x 10-18 J (1/nf2 - 1/nf2)


I know this has something to do with factoring -- but I forgot the rules that govern factoring. Like -2x-(-3x)=?

I should know this but it's been several years since my last math class.

Sponsored Links