The photoelectric effect consists of the emission of electrons (electric current) that occurs when light falls on a metal surface under certain conditions.

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.

<u>This is what Einstein proposed: </u>

Light behaves like a stream of particles called photons with an energy $E$ :

$E=\frac{hc}{\lambda}$ (1)

So, the energy $E$ of the incident photon must be equal to the sum of the Work function $\Phi$ of the metal and the kinetic energy $K$ of the photoelectron:

$E=\Phi+K$ (2)

Where $\Phi=6.94(10)^{-19} J$ 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:

<h3 /><h3>a) Maximum possible kinetic energy of the emitted electrons ( $K$ )</h3>

From (1) we can know the energy of one photon of 233 nm light:

$E=\frac{hc}{\lambda}$

Where:

$h=6.63(10)^{-34}J.s$ is the Planck constant

$\lambda=233 (10)^{-9} m$ is the wavelength

$c=3 (10)^{8} m/s$ is the speed of light

$E=\frac{(6.63(10)^{-34}J.s)(3 (10)^{8} m/s)}{3 (10)^{8} m/s}$ (3)

$E=8.53(10)^{-19} J$ (4) This is the energy of one 233 nm photon

Substituting (4) in (2):

$8.53(10)^{-19} J=6.94(10)^{-19} J+K$ (5)

Finding $K$ :

$K=1.59(10)^{-19} J$ (5) This is the maximum possible kinetic energy of the emitted electrons

<h3>b) Maximum number of electrons that can be freed by a burst of light whose total energy is $2 \mu J=2(10)^{-6} J$ </h3>

Since one photon of 233 nm is able to free at most one electron from the Titanium metal, we can calculate the following relation:

$\frac{E_{burst}}{E}$

Where $E_{burst}=2(10)^{-6} J$ is the energy of the burst of light

Hence:

$\frac{E_{burst}}{E}=\frac{2(10)^{-6} J}{8.53(10)^{-19} J}=2.34(10)^{12} electrons$ This is the maximum number of electrons that can be freed by the burst of light.

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3 years ago