Asked by Max89

Mercury vapour is used in as the dominant species which is excited in a traditional fluorescent light. The first and second excited states of Mercury occur at 254 and 185 nm.
Calculate the energies of each of these transitions?
What equation am i suppose to use to calculate this answer

Answers

Answered by drwls
Energy = h c/(wavelength)

h is Planck's constant, which must have been explained in your reading material.

c is the speed of light.

There are no AI answers yet. The ability to request AI answers is coming soon!

Related Questions