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## AP®︎/College Chemistry

### Course: AP®︎/College Chemistry>Unit 3

Lesson 12: Electronic transitions in spectroscopy

# Electronic transitions and energy

Electronic transitions occur in atoms and molecules due to the absorption or emission of electromagnetic radiation (typically UV or visible). The energy change associated with a transition is related to the frequency of the electromagnetic wave by Planck's equation, E = h𝜈. In turn, the frequency of the wave is related to its wavelength and the speed of light by the equation c = 𝜆𝜈. Created by Sal Khan.

## Want to join the conversation?

• One question on the video, when Sal Khan stated that there is no such thing as a 3 and a half shell, aren't there sp3 "hybrid" shells that form half-shells between one energy orbital and the next? • So when referring to electron shells, or energy levels, we are referring to the principal quantum number, n. Quantum mechanics states that energy, and by extension electrons, are quantized around the nucleus in distinct bands or levels. Meaning that electrons do no exist in between these energy levels. Principal quantum numbers are integer values, so n= 1,2,3, etc., but not fractional values like 1.5 or 3.25.

sp3, sp2, and sp refer to hybrid orbitals which compose an electron shell. Orbitals are the subunits of an electron shell. So there are 2sp3 and 3sp3 orbitals (again the number in front represents the principle quantum number), but not 2.5sp3 orbitals.

Hope that helps.
• Why did he change 486 nm to 486 * 10 ^-19? Doesn't that kind of complicate things?
(1 vote) • The reason Sal has to convert from nm to meters is because of the constants he is using in the bottom right of the screen. The values he's using for Planck's constant and the speed of light have units of J•s and m/s respectively. These units are using the base SI units which includes using meters (and not nanometers) for length, seconds for time, and kilograms for mass. And when we're doing math with scientific values and constants we have to have unit agreement, meaning they have to same units for all these properties or else we get wrong answers.

So even though 486 nm is simpler compared to 486•10^(-19) m, we have to have the wavelength in meters for the math to work out correctly.

If we really wanted to use 486 nm, we would just need new values for the constants which use nanometers instead of meters. However this usually makes things more complicated compared to simply converting the 486 nanometers to meters.

Hope that helps. 