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)

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  

$E=h.\nu$ (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:  

$E=h \nu$

Where:  

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

$\nu$ is the frequency  

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

$\nu=\frac{c}{\lambda}$ (3) Being $c=3(10)^{8}m/s$ is the speed of light in vacuum

$\nu=\frac{3(10)^{8} m/s}{400(10)^{-9}m}$ (4)

$\nu=7.5(10)^{14} Hz$ (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 $E$ (as in this situation) photoelectric effect cannot occur.