Given that, for Hydrogen, the ground state energy is -13.6 eV and the first excited state energy is -3.40 eV, and the second excited state energy is -1.51 eV I conclude that:(adsbygoogle = window.adsbygoogle || []).push({});

- if a 15 eV photon encounters a ground-state hydrogen atom, the photon can be absorbed by the hydrogen electron, freeing it from the nucleus & sending it away with kinetic energy of 1.4 eV.

- if a 10.2 eV photon encounters a ground-state hydrogen atom, it can be absorbed by the hydrogen electron, raising it to the first excited state.

- if a 8 eV photon encounters a ground-state hydrogen atom, it can either sail by (or through) unaffected at all, or it can collide elastically and be scattered but retain its initial energy (except for a tiny fraction to conserve momentum).

(Please correct me if I'm wrong about any of the above.)

But what happens if an 11 eV photon collides with such an atom? Does it

a. collide elastically just like the 8 eV photon?

or

b. give up 10.2 eV of its energy to raise the electron to the 1st excited state and continue on its way with its energy reduced to 0.8 eV?

(b) seems reasonable, but with this quantum stuff, obviously "reasonable" isn't good enough. So which is it?

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# Homework Help: More about photons

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