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Follow-Up Quiz -- Solution
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by Greg Jacobs
Woodberry Forest School
Woodberry, Virginia
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The diagram above shows two energy levels of a hypothetical atom.
- What wavelength photon will electrons emit as they make a transition from E2 to E1?
Answer: 520 nm
In addition to the wavelength calculated in (a), electrons that start in state E2 also emit photons of wavelength 413 nm.
- On the diagram above, draw and label another energy level E3 that would be consistent with the observation of the 413 nm emission.
Answer: 413 nm converts to a photon energy of 3.0 eV by 
. The electron in state E2 must drop by 3.0 eV, placing it in the state shown above with energy –4.8 eV.
- What other wavelength photon should be observed? Categorize this photon as ultraviolet, infrared, or visible.
Answer: Electrons that drop to the –4.2 eV state must then be able to drop to the –4.8 eV state. This transition would cause the emission of a photon with energy 0.6 eV. By
, the wavelength of this photon is 2,100 nm. This is a far longer wavelength than visible photons. In the electromagnetic spectrum, infrared photons have longer wavelengths than visible photons.
- Now, assume that electrons in state E2 absorb 100 nm photons. What will be the maximum speed of the electrons ejected from this atom?
Answer: 100 nm photons carry 12.4 eV of energy, by
. These electrons must use 1.8 eV to escape the atom, leaving 10.6 eV for kinetic energy.
Kinetic energy is given by 
; m is the electron's mass, and the kinetic energy is known, so students can find the speed v using algebra. The only subtlety is that this equation requires standard units to get a speed in meters per second. Students must convert the 10.6 eV kinetic energy to joules:
Now, solve for speed to get 1.9 × 107 m/s. This is a reasonable answer: it's less than the speed of light but far greater than everyday speeds. (You'll usually find electrons moving at 106 or 107 m/s.)
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