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

Step-by-step 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=(c)/(\lambda_(1))` (6)

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

Substituting (6) in (5):

`(hc)/(\lambda_(1))=6eV` (7)

Then finding
`\lambda_(1)`:

`\lambda_(1)=(hc)/(6eV )` (8)

`\lambda_(1)=((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)=(hc)/(\lambda_(2))` (12)

`E_(2)=((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