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

Answer:

a) $1.59(10)^{-19} J$

b) $2.34(10)^{12} electrons$

Explanation:

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.  

This is what Einstein proposed:

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:

a)  Maximum possible kinetic energy of the emitted electrons ($K$)

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

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

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.