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 [tex]E[/tex]:
[tex]E=h.f[/tex] (1)
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:
[tex]E=\Phi+K[/tex] (2)
Where [tex]\Phi[/tex] 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 [tex]\Phi=2eV[/tex] and [tex]K_{1}=4eV[/tex]
So, for the first light source of wavelength [tex]\lambda_{1}[/tex], and applying equation (2) we have:
[tex]E_{1}=2eV+4eV[/tex] (3)
[tex]E_{1}=6eV[/tex] (4)
Now, substituting (1) in (4):
[tex]h.f=6eV[/tex] (5)
Where:
[tex]h=4.136(10)^{-15}eV.s[/tex] is the Planck constant
[tex]f[/tex] is the frequency
Now, the frequency has an inverse relation with the wavelength
[tex]\lambda_{1}[/tex]:
[tex]f=\frac{c}{\lambda_{1}}[/tex] (6)
Where [tex]c=3(10)^{8}m/s[/tex] is the speed of light in vacuum
Substituting (6) in (5):
[tex]\frac{hc}{\lambda_{1}}=6eV[/tex] (7)
Then finding [tex]\lambda_{1}[/tex]:
[tex]\lambda_{1}=\frac{hc}{6eV } [/tex] (8)
[tex]\lambda_{1}=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{6eV}[/tex]
We obtain the wavelength of the first light suorce [tex]\lambda_{1}[/tex]:
[tex]\lambda_{1}=2.06(10)^{-7}m[/tex] (9)
Now, we are told the second light source [tex]\lambda_{2}[/tex] has the double the wavelength of the first:
[tex]\lambda_{2}=2\lambda_{1}=(2)(2.06(10)^{-7}m)[/tex] (10)
Then: [tex]\lambda_{2}=4.12(10)^{-7}m[/tex] (11)
Knowing this value we can find [tex]E_{2}[/tex]:
[tex]E_{2}=\frac{hc}{\lambda_{2}}[/tex] (12)
[tex]E_{2}=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{4.12(10)^{-7}m}[/tex] (12)
[tex]E_{2}=3.011eV[/tex] (13)
Knowing the value of [tex]E_{2}[/tex] and [tex]\lambda_{2}[/tex], and knowing we are working with the same work function, we can finally find the maximum kinetic energy [tex]K_{2}[/tex] for this wavelength:
[tex]E_{2}=\Phi+K_{2}[/tex] (14)
[tex]K_{2}=E_{2}-\Phi[/tex] (15)
[tex]K_{2}=3.011eV-2eV [/tex]
[tex]K_{2}=1.011 eV[/tex] This is the maximum kinetic energy for the second light source
