me too, i love your videos, literally a life saver. You should just rewrite the IB books because their explanations are nothing compared to yours honestly.
Love your videos, they are very organized and well explained. The only suggestion I have is whenever you introduce a new formula, writing down the definitions of each variable in a list form on the side will make your lecture even more clear.
Tord Thanks for the comment. Generally, I have been aiming to minimise writing in the videos but I agree that there are situations where a quick list would help a lot of viewers.
Thanks so much! I absolutely love how you explain stuff!! It makes it clear in my head as to what goes where and how I should solve problems on the topic!
In 9:43 the explanation you gave if there were more photons is opposite to what you gave in the previous section of comparing wave and photon. Am I right? Which one of these are right, I think that if there were more photons(brighter light), would emit the same number of electrons but with more energy. Please reply to clarify. Thank you.
Photon model wins. In this case, light behaves as photons. Each photon can free one electron, so more photons produces more electrons. The electrons are more energetic if the photons are more energetic, not if there are more photons.
12:55 Sir in this question: The intensity of the light is increased which means the light is brighter. We discussed that brighter light will produce more electrons, I get that. But the question says that the electrons are already being emitted, so that means the frequency is greater than the cut off frequency. Now, higher intensity means greater energy and furthermore, greater frequency, so shouldn't it be increase, increase?
so does the correct prediction, which is photon prediction tell that higher-frequency light means electrons are more energetic, but the amount is the same?
The work function is sort of like the ionization energy for a metal. Metals have a free sea of electrons. These electrons are not attached to any particular atom but to the overall lattice of positive ions. Ionization energy refers to the energy required to remove one electron from a neutral atom. The work function is the energy needed to remove one free sea electron from a metal.
me too, i love your videos, literally a life saver. You should just rewrite the IB books because their explanations are nothing compared to yours honestly.
Love your videos, they are very organized and well explained. The only suggestion I have is whenever you introduce a new formula, writing down the definitions of each variable in a list form on the side will make your lecture even more clear.
Tord Thanks for the comment. Generally, I have been aiming to minimise writing in the videos but I agree that there are situations where a quick list would help a lot of viewers.
Thanks so much! I absolutely love how you explain stuff!! It makes it clear in my head as to what goes where and how I should solve problems on the topic!
lord and saviour chris doner coming in clutch before mock 2 once again
Good luck!
This video was very helpful, but for the last on isn't 4.36eV-1.40eV = 2.96eV??? not 3.06eV
Thanks Kishan, annotation has been added.
In 9:43 the explanation you gave if there were more photons is opposite to what you gave in the previous section of comparing wave and photon. Am I right? Which one of these are right, I think that if there were more photons(brighter light), would emit the same number of electrons but with more energy. Please reply to clarify. Thank you.
Photon model wins. In this case, light behaves as photons. Each photon can free one electron, so more photons produces more electrons. The electrons are more energetic if the photons are more energetic, not if there are more photons.
@@donerphysics thanks sir!
12:55 Sir in this question: The intensity of the light is increased which means the light is brighter. We discussed that brighter light will produce more electrons, I get that. But the question says that the electrons are already being emitted, so that means the frequency is greater than the cut off frequency. Now, higher intensity means greater energy and furthermore, greater frequency, so shouldn't it be increase, increase?
Higher intensity will not change the frequency. The frequency is essentially a measure of the color of the light.
Just to clarify further, is cut off frequency the same as threshold frequency?
same same
so does the correct prediction, which is photon prediction tell that higher-frequency light means electrons are more energetic, but the amount is the same?
right.
This video is amazing! It helps me a lot. Thank you!
What is the difference between ionization energy and the work function (conceptually)?
The work function is sort of like the ionization energy for a metal. Metals have a free sea of electrons. These electrons are not attached to any particular atom but to the overall lattice of positive ions. Ionization energy refers to the energy required to remove one electron from a neutral atom. The work function is the energy needed to remove one free sea electron from a metal.
Isn't 4.36 Ev - 1.40 Ev = 2.96 Ev for the final question?
This mistake has been pointed out in the comments before. An annotation was made but I don't think the annotations show up anymore.
Bless your videos
Great video
Legendary! As always :)
Thanks again!