@@gabriellita7236 con his intendevo l’inglese del mio professore. La lezione era in inglese e nella verifica la risposta sarebbe stata da dare in inglese
Oh my gosh. In under 10 minutes you have explained more than my teacher has done in about 12 lessons, and you made it even more understandable. Thank you so much
Someone in my class showed me the textbook about this topic and I was sitting there like "what???" but you made it so easy to understand, keep doing what your doing!!
broooo, you’re the man 👍🏻👍🏻👍🏻👍🏻 my teacher waffled on and on for many lessons, it would’ve been much easier if he just showed us this video. I subbed and Liked
Thank you! I know this video is 5 year old, but it's still coming in handy. Got a physics exam in 2 and a half weeks and this is great consolidation on the photoelectric effect :)
body decided to get the flu during the week so i missed the lessons where my teacher explained this stuff. this really helped me, thank you! will come back for my end of years :'-)
I liked it at 0:01, as every vídeo of yours is amazing 😉 Thank you so much, I am preparing for PAT and this is sooo helpful. It’s really calming to discover you have a video on what I need to learn 😁
Really helpful, concise, very understandable. Thank you very much for this lesson. You are even better than the PhD teacher of my college !!! in terms of explanation.
awesome job...thee BEST i have seen...very well explained in simple, and clear demonstration..thank you SO much..now i can REALLY enjoy all my higher mathematics books...!!
Given that you have frequency a and b and that a and b are greater than the threshold frequency and b is greater than a. If you read current x on the ampmeter with freq a, what will(or will not) happen with the current when you switch the incoming light to freq b? Intensity is to remain constant, i.e the machine that’s generating the photons is always running on say 10W=10J/s with both freq a and b. (assume straight line photon beam) Thanks for the great video!
Hi! Thanks, great video! I have a question though. The curcuit you illustrated at 2:33, why did you change the poles? I guess it works both ways, but wouldn't the energy threshhold for the electrons to be jumping over to the other metal plate be much lower before you changed the poles? I mean that it would be an electric field between the two poles. Does your explanation say that the electrons travel in the same direction as the current, if the photoelectric effect is "strong enough"? Best regards
you did. the quanta itself is the wave its the spin properties that drive the wave function. also the capacitance of electrons is spectrum dependant. 8:1 infrared photons to visible spectrum and this is the reason its colour not carbon for climate change
I think that an electron is a positively charged ion, which should be emitted from the metal plate(cathode) towards the collector (Anode). Also, I think that negatively charged ions are attracted to positively charged ions. Why are the electrons emitted from the anode instead of from the cathode which emits electrons in your diagram? Also, do we have an electron that is a positively charged ion? I need clarifications, sir. Thank you.
Waves overlap with each other when you add multiple inphase waves. That should give more energy to electron but that doesn't happen in photoelectric effect.
