Question

If the work function of a material is such that red light of wavelength 700 nm just barely initiates the photoelectric effect, what must the maximum kinetic energy of ejected electrons be when violet light of wavelength 400 nm illuminates the material?Express your answer with the appropriate units.Kmax = J

Answer 1

Answer: $2.13(10)^{-19} J$

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  $E$

$E=h.f$ (1)

Where:

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

$f$ is the frequency

Now, the frequency has an inverse relation with the wavelength $\lambda$:  

$f=\frac{c}{\lambda}$ (2)  

Where $c=3(10)^{8}m/s$ is the speed of light in vacuum  and $\lambda=400nm=400(10)^{-9}m$ is the wavelength of the absorbed photons in the photoelectric effect.

Substituting (2) in (1):

$E=\frac{h.c}{\lambda}$ (3)

So, the energy $E$ of the incident photon must be equal to the sum of the Work function $\Phi$ of the metal and the maximum kinetic energy $K_{max}$ of the photoelectron:  

$E=\Phi+K_{max}$ (4)  

Rewriting to find $K_{max}$:

$K_{max}=E-\Phi$ (5)

Where $\Phi$ is the minimum amount of energy required to induce the photoemission of electrons from the surface of a metal, and its value depends on the metal:

$\Phi=h.f_{o}=\frac{h.c}{\lambda_{o}}$ (6)

Being $\lambda_{o}=700nm=700(10)^{-9}m$ the threshold wavelength (the minimum wavelength needed to initiate the photoelectric effect)

Substituting (3) and (6) in (5):  

$K_{max}=\frac{h.c}{\lambda}-\frac{h.c}{\lambda_{o}}$

$K_{max}=h.c(\frac{1}{\lambda}-\frac{1}{\lambda_{o}})$ (7)

Substituting the known values:

$K_{max}=(6.63(10)^{-34}J.s)(3(10)^{8}m/s)(\frac{1}{400(10)^{-9}m}-\frac{1}{700(10)^{-9}m})$

$K_{max}=2.13(10)^{-19} J$ >>>>>This is the maximum kinetic energy that ejected electrons must have when violet light illuminates the material