Respuesta :
Answer:
a) [tex]1.59(10)^{-19} J[/tex]
b) [tex]2.34(10)^{12} electrons[/tex]
Explanation:
The photoelectric effect consists of the emission of electrons (electric current) that occurs when light falls on a metal surface under certain conditions. Â
If the light is a stream of photons and each of them has energy, this energy is able to pull an electron out of the crystalline lattice of the metal and communicate, in addition, a kinetic energy. Â
This is what Einstein proposed:
Light behaves like a stream of particles called photons with an energy  [tex]E[/tex]:
[tex]E=\frac{hc}{\lambda}[/tex] (1) Â
So, the energy [tex]E[/tex] of the incident photon must be equal to the sum of the Work function [tex]\Phi[/tex] of the metal and the kinetic energy [tex]K[/tex] of the photoelectron: Â
[tex]E=\Phi+K[/tex] (2) Â
Where [tex]\Phi=6.94(10)^{-19} J[/tex] is the minimum amount of energy required to induce the photoemission of electrons from the surface of Titanium metal.
Knowing this, let's begin with the answers:
a) Â Maximum possible kinetic energy of the emitted electrons ([tex]K[/tex])
From (1) we can know the energy of one photon of 233 nm light:
[tex]E=\frac{hc}{\lambda}[/tex]
Where:
[tex]h=6.63(10)^{-34}J.s[/tex] is the Planck constant Â
[tex]\lambda=233 (10)^{-9} m[/tex] is the wavelength
[tex]c=3 (10)^{8} m/s[/tex] is the speed of light
[tex]E=\frac{(6.63(10)^{-34}J.s)(3 (10)^{8} m/s)}{3 (10)^{8} m/s}[/tex] (3)
[tex]E=8.53(10)^{-19} J[/tex] (4) This is the energy of one 233 nm photon
Substituting (4) in (2):
[tex]8.53(10)^{-19} J=6.94(10)^{-19} J+K[/tex] (5) Â
Finding [tex]K[/tex]:
[tex]K=1.59(10)^{-19} J[/tex] (5) Â This is the maximum possible kinetic energy of the emitted electrons
b) Maximum number of electrons that can be freed by a burst of light whose total energy is [tex]2 \mu J=2(10)^{-6} J[/tex]
Since one photon of 233 nm is able to free at most one electron from the Titanium metal, we can calculate the following relation:
[tex]\frac{E_{burst}}{E}[/tex]
Where [tex]E_{burst}=2(10)^{-6} J[/tex] is the energy of the burst of light
Hence:
[tex]\frac{E_{burst}}{E}=\frac{2(10)^{-6} J}{8.53(10)^{-19} J}=2.34(10)^{12} electrons[/tex] This is the maximum number of electrons that can be freed by the burst of light.