MCAT Chemistry Question — Photon
- Jun 26, 2017
- MCAT Question of the Day
In a phenomenon called the photoelectric effect, a photon striking the surface of a metal may be absorbed and lead to the emission of a single electron. The energy of the emitted electron is described by E = φ-hf, where φ is the work function, h is Planck’s constant (6.626*10^-34 J*s or 4.136 * 10^-15 eV*s) and f the frequency of incident light. If 3 photons of 200 nm wavelength are incident on a filament of tungsten (φ = 4.5 eV), how many electrons are emitted?
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Answer: A. 0 electrons will be emitted.
Kinetic energy is a scalar quantity that cannot be less than 0. The reader is left to make the inference that if the energy of emission is less than zero, no electron would be emitted.
We can verify what the maximum wavelength of a photon must be to stimulate emission from tungsten by rearranging E = hf-φ to solve first for frequency and then for wavelength:
E+φ = hf
(E+φ)/h = f
Now recall that c=f*λ and therefore f = λ/c and substitute, to get:
(E+φ)/h = λ/c
c*(E+φ)/h = λ
Now let’s plug in our numbers (E = 0 for the maximum wavelength of the incident photon for emission of an electron):
3*10^8 m/s * 4.5 eV / (4.136 * 10^-15 eV/s) = 1.088 * 10^-7 m
1.088 * 10^-7 m = 108.8 nm
Notice also that the energy for the emission of an electron must be provided by a single photon, so it is irrelevant that the three incident photons together will exceed the work function (φ).
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