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We're going over quantum basics in chemistry right now and I'm very confused.

Electrons can only accept in discreet quanta to move up an energy level, right? And they reflect other forms of light that don't supply energy in their specific quanta, right? And flame is just infrared electromagnetic radiation, right?

Then when heated by flame, why do the electrons move up to a next energy level? Shouldn't they need a certain wavelength to move up - one that isn't provided by the flame, given the emission for the atom is in the visible light range?

Qmechanic
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Pearson Frank
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  • https://physics.stackexchange.com/questions/460801/why-do-atoms-iron-eg-glow-with-all-frequencies-of-light-when-exposed-to-enough – BowlOfRed Oct 23 '19 at 05:06

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There's not really any such thing as heating an individual atom. When you heat a gas like in a candle flame, the heat is the random motion of all the different atoms. The randomness is what makes us call it heat. If there's only one atom, there's no way to say if its motion is random or not.

And they reflect other forms of light that don't supply energy in their specific quanta, right?

Or they just transmit it (act transparent to it).

Then when heated by flame, why do the electrons move up to a next energy level? Shouldn't they need a certain wavelength to move up - one that isn't provided by the flame, given the emission for the atom is in the visible light range?

The atoms are being excited by collisions with other atoms. They're not being excited by absorbing light.

  • This makes a little more sense. But how does the excitement caused by collision guarantee the right energy needed to move the electrons up a little? – Pearson Frank Oct 23 '19 at 12:57
  • @PearsonFrank: As the atoms are coming in toward their collision, they have some total amount of kinetic energy K available. This energy doesn't have to be fine-tuned to equal the excitation energy desired. This is because this is a scattering process, not absorption. After the collision, they have some reduced kinetic energy K'. Regardless of what random value K has, as long as K is big enough there is always some possible value of K' such that $K-K'$ is equal to the excitation energy you're talking about. –  Oct 23 '19 at 14:29