Question

What is the maximum kinetic energy (in eV) of a photoelectron when a surface, whose work function is 5.0 eV, is illuminated by photons whose wavelength is 400 nm?

Answer 1

Answer: -1.898 ev (this means the photoelectric effect cannot occur)

Step-by-step 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=h.\\u` (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 surface.

In this case
`\Phi=5 eV`

So, applying equation (1) in this problem:

`E=h \\u`

Where:

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

`\\u` is the frequency

Now, the frequency has an inverse relation with the wavelength
`\lambda=400(10)^(-9)m`:

`\\u=(c)/(\lambda)` (3) Being
`c=3(10)^(8)m/s` is the speed of light in vacuum

`\\u=(3(10)^(8) m/s)/(400(10)^(-9)m)` (4)

`\\u=7.5(10)^(14) Hz` (5)

Substituting (5) in (1):

`E=(4.136(10)^(-15)eV.s)(7.5(10)^(14) Hz)` (6)

`E=3.102 eV` (7)

Now, substituting (7) in (2):

`3.102 eV=5 eV+K` (8)

Finding
`K`:

`K=-1.898 eV` (9) Since a negative kinetic energy is not physically possible, the only explanation is: the conditions for a photoelectric effect were not met, hence the photoelectric effect cannot occur.

To understand it better:

The main condition for the occurrence of the photoelectric effect is that the energy incident photon
`E` must be greater than the work function
`\Phi`

`E>\Phi`

If
`E<\Phi` (as in this situation) photoelectric effect cannot occur.