Blue light waves have more energy than red light waves. which of these situations most likely produces a photoelectric effect? you may assume that the same metal is used in each situation. check all that apply. shining dim red light on a strip of metal shining bright red light on a strip of metal shining dim blue light on a strip of metal shining bright blue light on a strip of metal

The correct options are:

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:

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.

the answer is definitely c.