When red light shines on a piece of metal, no electrons are released. When the red light is slowly changed to shorter-wavelength light (basically progressing through the rainbow), nothing happens until yellow light shines on the metal, at which point electrons are released from the metal. If this metal is replaced with a metal having a higher work function, which light would have the best chance of releasing electrons from the metal? If this metal is replaced with a metal having a higher work function, which light would have the best chance of releasing electrons from the metal? And please explain why and list any equations if there are any. Thanks!

A. Red.
B. Blue.
C. Yellow would still work fine.
D. We need to know more about the metals involved.

This is an exercise of photoelectric effect where electrons are removed from the metal by the incident radiation, the explanation of this phenomenon was given by Einstein, they assume that the light is formed by quanta and these collide with the electrons promoting their transfer; The equation that describes the process is

[tex]K_{max}[/tex] = h f - Φ

Where[tex]K_{max}[/tex]is the kinetic energy, maximize the electrons started, h the plan constant, f the frequency and Φ the work function of the metal (link energy)

The radiation with the lowest energy that an electron can start

h f = Φ

Let's analyze the energy of the light rays with the wavelength

c = λ f

f = c / λ

λ red = 700 10⁻⁹ m

f = 3 108/700 10⁻⁹

f₁ = 4.29 10¹⁴ Hz

Yellow λ = 600 10⁻⁹ m

f₂ = 3 108/600 10⁻⁹

f₂ = 5 10¹⁴ Hz

λ blue = 450 10⁻⁹ m

f₃ = 6.67 10¹⁴ Hz

In the initial metal with a work function fi1 yellow light was needed, to start the electrons, when analyzing the initial equation if the work function increases Φ₂> Φ₁ more energy is needed to take out the electrons from the atom (E₂>E₁)

light of greater energy (frequency) than yellow is needed so that light is needed to go BLUE