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 = hfφ, 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 is 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?
A. 0
B. 1
C. 2
D. 3
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Answer: D. 3 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 is to stimulate emission from tungsten by setting E = 0 and rearranging E = hfφ to solve for
frequency and then wavelength:
E+φ = hf
(E+φ)/h = f
Now recall that c = f*λ and therefore f = c/λ and substitute, to get:
(E+φ)/h = c/λ
λ*(E+φ)/h = c
λ = c*h/(E+φ)
Now let’s plug in our numbers (remember E = 0)
3*10^8 m/s * (4.136 * 10^15 eV/s) / 4.5 eV = 2.757 * 10^7 m
2.757 * 10^7 m = 275.7 nm
The maximum wavelength of a photon to stimulate electron emission from tungsten is therefore 275.7 nm. Since the 3 incident photons are each 200 nm,
all less than the maximum wavelength, each photon will result in the emission of one electron.
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