Exercise PhysicsWiki: Difference between revisions
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<math>\frac{-\hbar^2}{2m}\nabla^2\psi(r) + V(r)\psi(r) = E\psi(r)</math> (Brandon Bryant) | <math>\frac{-\hbar^2}{2m}\nabla^2\psi(r) + V(r)\psi(r) = E\psi(r)</math> (Brandon Bryant) | ||
<math>E = mc^2</math> Cheyvonne Perry |
Revision as of 10:38, 31 August 2009
$$a=b$$
(posted by TerriC, Group 2)
(Zach McDargh)
(posted by KimberlyWynne 19:05, 29 August 2009 (EDT))
(Sandy Simmons)
(Steve Honeywell)
(Andrew Wray)
(Ryan Taylor)
(August Larson)
(Brandon Bryant)
Cheyvonne Perry