What is the maximum number of electrons that can be freed by a burst of light whose total energy is 2.00 µJ.

Question

Titanium metal requires a photon with a minimum energy of 6.94×10−19J to emit electrons.

 

 

If titanium is irradiated with light of 233 nm, what is the maximum possible kinetic energy of the emitted electrons?

What is the maximum number of electrons that can be freed by a burst of light whose total energy is 2.00 μJ.

 

 

What is the maximum number of electrons that can be freed by a a burst of light whose total energy is 2.00 μJ.

Answer 1

The energy of each photon can be obtained using Planck's equation:  E = hν = (hc)/λ = (6.626 x 10^-34 J s)(2.9979 x 10⁸ m/s) / (233 x 10^-9 m) = 8.525 x 10^-19 J

 

The maximum kinetic energy for each freed electron will be 8.525 x 10^-19 J - 6.94 x 10^-19 J = 1.585 x 10^-19 J

 

The number of 233 nm photons in the 2.00 x 10^-6 J burst of light will be 2.00 x 10^-6 J / 8.525 x 10^-19 J = 2.346 x 10¹² photons.  Each photon can free at most one electron, so the maximum number of freed electrons will be 2.346 x 10¹² electrons.