Question

A light source of wavelength λ illuminates a metal with a work function (a.k.a., binding energy) of BE=2.00 eV and ejects electrons with a maximum KE=4.00 eV. A second light source with double the wavelength of the first ejects photoelectrons with what maximum kinetic energy?

Answer 1

Answer: 1.011 eV

Explanation:

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

If we consider the light as 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)  

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 $\Phi$ 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.  

In this case $\Phi=2eV$  and $K_{1}=4eV$

So, for the first light source of wavelength $\lambda_{1}$, and  applying equation (2) we have:

$E_{1}=2eV+4eV$   (3)  

$E_{1}=6eV$   (4)  

Now, substituting (1) in (4):  

$h.f=6eV$ (5)  

Where:  

$h=4.136(10)^{-15}eV.s$ is the Planck constant

$f$ is the frequency  

Now, the frequency has an inverse relation with the wavelength

$\lambda_{1}$:  

$f=\frac{c}{\lambda_{1}}$ (6)  

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

Substituting (6) in (5):  

$\frac{hc}{\lambda_{1}}=6eV$ (7)  

Then finding $\lambda_{1}$:  

$\lambda_{1}=\frac{hc}{6eV }$ (8)  

$\lambda_{1}=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{6eV}$  

We obtain the wavelength of the first light suorce $\lambda_{1}$:  

$\lambda_{1}=2.06(10)^{-7}m$   (9)

Now, we are told the second light source $\lambda_{2}$  has the double the wavelength of the first:

$\lambda_{2}=2\lambda_{1}=(2)(2.06(10)^{-7}m)$   (10)

Then: $\lambda_{2}=4.12(10)^{-7}m$   (11)

Knowing this value we can find $E_{2}$:

$E_{2}=\frac{hc}{\lambda_{2}}$   (12)

$E_{2}=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{4.12(10)^{-7}m}$   (12)

$E_{2}=3.011eV$   (13)

Knowing the value of $E_{2}$ and $\lambda_{2}$, and knowing we are working with the same work function, we can finally find the maximum kinetic energy $K_{2}$ for this wavelength:

$E_{2}=\Phi+K_{2}$ (14)  

$K_{2}=E_{2}-\Phi$ (15)  

$K_{2}=3.011eV-2eV$  

$K_{2}=1.011 eV$  This is the maximum kinetic energy for the second light source