The work function of aluminum is 4⋅2 eV. If two photons each of energy 2⋅5 eV are incident on its surface, will the emission of electrons take place? Justify your answer.
By using Einstein's photoelectric equation is, K. E. = hv - Φ or hv = Φ + K. E. we can say that the emission of electrons takes place only if the incident energy is greater than the work function of the material.
In the given case, total incident energy =2 × 2.5 eV = 5 eV
The work function of aluminum, Φ = 4.2 eV
∵ Incident energy is greater than the work function of the material, hence the emission of electrons will take place.
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