Answer:
The correct options are:
- shining dim blue light on a strip of metal, and
- shining bright blue light on a strip of metal
Step-by-step explanation:
The photoelectric effect is the emission of electrons from a metallic surface when light of determined frequency shines over the metallic surface.
The electrons are ejected from the metal surface when the frequency of the incident light beam reaches a minimum frequency (threshold).
The frequency of light is related to the color of light (not with the intensity).
The red light is at the lower frequency end and the violet light at the higher frequency end of the spectrum of visible light. That condition is considered in the given statement when it tells that "Blue light waves have more energy than red light waves" , since the frequency is directly related to the energy, by the Planck equation:
Energy of a photon (light) = Planck constant × frequency.
The intensity of the light (if the light is dim or bright) does not influence the photoelectric effect.
Hence, you conclude that the light with more energy (the blue light) either dime or bright will most likely produce the photoelectric effect. That is why the correct options are:
- shining dim blue light on a strip of metal, and
- shining bright blue light on a strip of metal