Photoelectric effect, and sunlight.

AI Thread Summary
The discussion centers on whether sunlight has sufficient intensity to eject electrons from metal atoms through the photoelectric effect. It clarifies that while sunlight has a high enough frequency, visible light alone cannot knock out electrons; ultraviolet light is typically required for this effect. The intensity of light influences the number of electrons emitted but not the ability to eject them, which depends on frequency. Additionally, atmospheric conditions block much of the UV radiation, limiting its effectiveness. Overall, the conversation highlights the importance of frequency over intensity in the context of the photoelectric effect.
kcajrenreb
Messages
49
Reaction score
0
Hello, I apologize if this question makes no sense, or is stupid, but I would just like to clarify something.

Is the light that comes from the sun a high enough intensity to "knock" any electrons out of the atoms in a metal? I know it is a high enough frequency, and I think the intensity is high enough, but I'd just like to make sure.

Thank you very much, and again I apologize if my question made little sense.
 
Physics news on Phys.org
The photoelectric effect is independent of intensity. A single photon can knock an electron out of metal as long as the frequency is high enough, aka it has enough energy. Intensity merely causes MORE electrons to be knocked out per unit of time.
 
kcajrenreb said:
Hello, I apologize if this question makes no sense, or is stupid, but I would just like to clarify something.

Is the light that comes from the sun a high enough intensity to "knock" any electrons out of the atoms in a metal? I know it is a high enough frequency, and I think the intensity is high enough, but I'd just like to make sure.

Thank you very much, and again I apologize if my question made little sense.

Visible light cant.
X-ray i think can knock out an electron of almost everything.
The problem is if they can get to your metal. most of UV and higher get blocked by the atmosphere.
 
Really? I thought that blue light on the visible spectrum was a high enough frequency to "knock" the electrons out of metal, I guess I thought wrong. Thanks for your answers both of you.
 
kcajrenreb said:
Really? I thought that blue light on the visible spectrum was a high enough frequency to "knock" the electrons out of metal, I guess I thought wrong. Thanks for your answers both of you.

From wikipedia:
When a surface is exposed to electromagnetic radiation above a certain threshold frequency (typically visible light for alkali metals, near ultraviolet for other metals, and extreme ultraviolet for non-metals), the radiation is absorbed and electrons are emitted. Light, and especially ultra-violet light, discharges negatively electrified bodies with the production of rays of the same nature as cathode rays.

The article is here:http://en.wikipedia.org/wiki/Photoelectric_effect
 
Thread 'Griffith, Electrodynamics, 4th Edition, Example 4.8. (Second part)'
I am reading the Griffith, Electrodynamics book, 4th edition, Example 4.8. I want to understand some issues more correctly. It's a little bit difficult to understand now. > Example 4.8. Suppose the entire region below the plane ##z=0## in Fig. 4.28 is filled with uniform linear dielectric material of susceptibility ##\chi_e##. Calculate the force on a point charge ##q## situated a distance ##d## above the origin. In the page 196, in the first paragraph, the author argues as follows ...
Thread 'Inducing EMF Through a Coil: Understanding Flux'
Thank you for reading my post. I can understand why a change in magnetic flux through a conducting surface would induce an emf, but how does this work when inducing an emf through a coil? How does the flux through the empty space between the wires have an effect on the electrons in the wire itself? In the image below is a coil with a magnetic field going through the space between the wires but not necessarily through the wires themselves. Thank you.
Back
Top