Photoelectric effect | Electronic structure of atoms | Chemistry | Khan Academy
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- Опубликовано: 14 окт 2024
- Explaining the photoelectric effect using wave-particle duality, the work function of a metal, and how to calculate the velocity of a photoelectron. Created by Jay.
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Funny how my chem teacher has been teaching this for 2 weeks now but i understood it in a 10 min video. Thankyou so much!!
When Khan academy teaches you more in one night than your physics teacher did in a whole semester...
Classic stich-up
William Fitzmaurice you're fuckin' goddamn right
William Fitzmaurice because we don’t pay attention in class and are forced to finish the whole portion in one night....and of course we don’t know crap that was taught in class so we refer to khan academy
In 2 years bro
so fricking true mate
Plz dont lie in My Empire
Amazing. This is a refuge for me after reviewing physics books that make you doubt your mental health rather than understanding the basic concept.
My Physics teachers are terrible. You sir are wonderful.
@Pranav Chopkar Nothing was wrong with the comment. Ironically, you’re the wrong one.
Dude deleted his comment
And if u pay even ten percent of attention paid while listening to khan academy in Physics class, , ur comments about ur teacher will be different.......its pathetic to see how u blame ur teachers for ur negligence.....n u r terrible student of Physics
@@syedashafiah2043 Hi Syeda, interesting of you to make a comment about the quality of my teachers despite never having been taught by them! 6 years on from my original comment, I still believe that they were poor teachers. I think the fact that I was on the internet searching for alternative methods of learning is evidence enough that I was paying attention which, in the end, paid off as I got a very high A (borderline A*) in physics at A level after moving colleges and getting better teachers. I'm now in the final stages of my master's degree in chemistry, for which I'm on-track to achieve a first-class award, and I'm hoping to pursue a PhD. I'm proud of the progress I've made academically, and I hope you see how shallow-minded your comment appears.
Your English teacher needs to get fired
I had to look everywhere to find the answer to my question, this really helped, thanks a lot!
Hpppppp
thank you so much. It was immensely intelligible!!!!!
we cannot use the formula E=MC^2 for calculating energy of each photon as there is different different formula for both momentum and energy as told in one of the video of sal khan at his site. as asked by viewers
Wow, wonderful. Thanks a lot
I actually understood this
thank youuuu
Thank you so much for this simple explanation that opens doors.
Why is this in Organic Chemistry?
+Blake Bullwinkel no it's not
+Blake Bullwinkel i have 3 of the most popular ochem textbooks. none of them include this
+violinsheets He means the channel
+NatSPlay oh but who cares? this channel has alot of random videos that aren't really relevant to ochem...at least they exist...
Blake Bullwinkel because swagggg
You did not mention the threshold frequency!
The work function is the threshold freqency*h
EDIT:h=6.626E-34J*s
Bre Donovan What you wrote there is Plank's constant and has nothing to do with this. The work function is metal specific and the threshold frequency is a FREQUENCY at which the effect would be able to take place.
+Dennis Premoli
please refer to this website:
chemistry.about.com/od/electronicstructure/a/photoelectric-effect.htm
Specifically where it defines what the work function is: "W is the work function, which is the minimum energy required to remove an electron from the surface of a given metal: hv_0".
EDIT: also if you choose to read what I wrote, I specifically mentioned the threshold FREQUENCY.
Bre Donovan Thank you but I had my baccalaureate in physics yesterday so...
I wasn't questioning your knowledge of Physics? You said I didn't mention the threshold frequency, but I had. Go back and read my first comment in its entirety.
Thank you!
Can we really use the classical mechanics EK=0.5mv^2 when there are relativistic effects at play? Wouldn't the kinetic energy equal "(mc^2)/sqrt(1-(v^2/c^2)) - mc^2" to account for the relativistic mass of the particles? Or is the effect negligible?
7:42 How did you know the mass of the electron? Is it something you just have to memorize?
difference between work function and ionisation energy?
One is from the solid
Another is from the gas!!
i literally love u wow thanks
I think you converted the nm to m wrong. 525 nm should be 525×10 ^-7 or 5.25 × 10^-9 . If I'm correct the answer is 3.78 × 10^-21 J not 3.78×10^-19 J
Which software you used?
The experiment overall is a capacitor tuned to specific frequencies. It's a band pass circuit hence why depending on the metal only certain color of light can activate the circuit.
Sir you are a genius. Gratitude from Bangladesh :D
Wow really good explanation thank you very much Sir
so if a photon frees an electron from the object would the object eventually lose all its electrons?
when an electron leaves ,in place of the electrons holes are created and result into positive current (opposite flow) and this has resulted into the discovery of semiconductor junctions and revolutionised the electronic industry in 1900's
nope, they will eventually fall back and emmit light for example
thats how the needed energy for the excitation of an electron in a spesific metal is being measured
does this result in the metal taking on a positive charge? do metals left in sunlight take on a positive charge? or is the photoelectric effect only observed in negatively charged metals, until they reach a neutral charge? do the photons continue off after striking the electrons?
Garl Jo'ens They gets emitted out of the metal surface and once they loose all their received energy, they return to the metal surface. So metals do not gain positive charge
Solar panels use the same mechanism.
What happens to the energy that we are calling work function?
False not logic statements from the beginning. 1 - there is no "KICK COLLISION WITH ELECTRON" , but absorbtion of photons by nucleus of the metal atom and EMITION of the electron caused by ATOM SPACE EXPANSION (Bohr quantum jump and Einstein fabric stretching by matter=energy). 2- look as well on your logic "massless photon kicks electron" isn't that a pure paradox ?
very helpful
That's my request!
great explanations, except the units of the last equation aren't explained at all. How does taking the root of V^2 = 7.69 *10^10*J*kg^-1 leave us with an answer with m/s?
Thank you a lot
I love Khan academy you make my day. I was having a hard time understanding
Why is this video so quiet?
Because it is educational
So work function is the same thing as threshold frequency right?
If the photon is massless, how does it knock the electron away? I'm still pretty new to this and I dont really understand that part
One word: momentum.
Even photons have momentum which is how solar sails in space are affected by light.
Momentum of a wave Plank's constant (h) divided by wavelength (λ)
what ll happen to photon energy when given frequency is less than threshold frequency??
It will go bing bong
thanks a lot sir.you are doing a gr8 job
Would the equation for this type of problem that will give you the answer in one calculation be:
V=sqrt((2(h*c-(lambda)*w))/m*lambda)?
Why some of energy will not be used to increase the temperature of the metal ?
They are talking about ideal situations. YOU can see some numerical where only some percent of energy is used as photoelectric emission other percent is heat.
Prakash Doshi black body radiation 100% efficient which is assumed if not stated otherwise
Solar panels use the same mechanism, they do get hot and only a part of the energy is harnessed as electricity.
can we just use "hc" as a constant in all of the problems that require it, since both "h" and "c" are constant, the only varying term there would be the wavelenght
Thank you this was so helpful
where did the mass came from? is it suppose to be in the problem?
i dont get it
after the electron is blasted off the metal by the photon what would happen to the metal(assuming a lot of photoelectrons are blasted)
plz do reply thx a lot in advance:)
I'm guessing it turns into a bunch of nuclei, or maybe if all protons decay into neutrons, possibly, it could turn into a mini neutron star.
I may be wrong but after constant bombardment of photons the metal will turn to positive ion and it will increase the work function of metal so as to oppose high attraction between high positive nuclei and small no of electron. So it is better to change metal after some time or it will be harder to eject electrons with passage of time. Maximum work function will be when there will be last electron in pool of positive nuclei in metal. Hope it is helpful!!!!
You can use the free electrons to run electrical appliances as you use solar panels.
Hi, I'm reading Beiser's modern physics. The kinetic energy of the photoelectron in that equation is actually referred to as the maximum kinetic energy of the photoelectron. So apparently a photoelectron can have a range of kinetic energies from the extraction to the maximum energy. Can anyone explain?
Reading further on, the kinetic energy is determined by the wavelength of the diffracted ray, which is determined by the angle of diffraction.
The sig figs give me a hemorrhage, going to 4 sig figs for constant, when max in the qn is 3 and then giving the answer with 2 when the min in the question is 3, great exam technique
There is an inaccurate information here. E0, the energy needed to free the electron from the metal, is equal to the work function ONLY for the most energetic photoelectrons, not for all photoelectrons. Photoelectrons are emitted with a range of energy, because not only do they have to overcome the work function, but also the binding energy. Only the conduction electrons that are at the Fermi surface (i.e. at the top of the conduction band), will be emitted with only the work function to overcome. All other electrons will have to overcome the work function and the binding energy.
This is why in a photoelectric effect experiments, we try to find the reverse potential that will stop the MOST ENERGETIC photoelectrons.
Very well presented. I was struggling with the concept until I saw the calculation using the larger wavelength and it then became crystal clear. Thanks!
so if a photon collides with an electron, is not really a collision beause the energy is absorbed right?
try a "work function" metal plate, with strips of voltaic pile strips, to make a non-electrolytic macro solar cell
Thank You
I wish people would stop commenting “I’m here because my professor sucks”. Yeah, we know, that’s why all of us are here.
Do you really think someone who understands the lecture perfectly would waste 10 minutes watching this instead of bathing in an ivory bathtub or whatever it is people with decent teachers do?
great . thanks 😊
Hi parya , are you iranian?
👍💯
good
Khan Academy=Albert Einstein no.2
I love the idea on how einstein did this and earned a nobell price and we now have to learn this kind of stuff in the university
It does NOT prove light is a particle. What if the atom itself can only accep discrete energies, rather than light itself.
what if (energy of electron)=work function
will the electron will be free with no velocity??
yes, in that case v=0, but we get free electron
@@huzaifaabedeen7119 thanks buddy
''.. and the darkness comprehend it not ''
Where do you get the mass ? It's not situated inside the sentence
The mass of an electron is a constant, it is always 9.109X10^-31
+wilson fink 9.109*10^-31 kg
Ask an astronomer "How are photons created or destroyed", talks about quantum jumps.
Q. The only interactions a photon has are to be created or destroyed.
My idea is that the photon pops up as an electron positron pair that power the quantum leap or releases a photon.
That means that the leaping electron interacts with the electron positron pair. One electron and positron annihilate which powers the left over electron to jump.
That means that in space time photons are always electron positron pairs.
This one's for my Father (Yoda). The Reason the effect Occurs is because the Electrons Surrounding the Atom aquire this Aether (High Surface Area Light Push to Low Volume Pull). Them electrons then Eject because of the Light Push is higher than Volume Pull. In the Plum pudding model. Bigger plums are generally in the centre and Smaller ones on the outside. However a few small Plums can be in the middle and a few Large Plums on the outside.
How to increase Solar panels efficiency. Dope the Junction with Large attom Mettals. Large like (Uranium) and High Conductivity like Gold.
- يبدو أحيانًا أن الضوء يعمل كموجة، لتعمل كجسيم. وأحيانًا يبدو الضوء ومثال على ذلك ، سيكون التأثير الكهروضوئي ، كما وصفه أينشتاين. لنفترض أن لديك قطعة من المعدن ، ونعلم أن المعدن به إلكترونات. ارسم إلكترونًا واحدًا هنا ، سأذهب إلى الأمام و وهذا الإلكترون مرتبط بالمعدن لأنه منجذب إلى الشحنات الموجبة في النواة. إذا سلطت الضوء على المعدن ، لذا فإن النوع المناسب من الضوء بالنوع الصحيح من التردد ، من تلك الإلكترونات فضفاضة ، يمكنك في الواقع أن تطرق بعضها من الإلكترونات لتدفق. الذي يسبب التيار لذلك هذا نوع من الاصطدام بين جسيمين ، كجسيم. إذا فكرنا في الضوء لذا سأرسم جسيمًا من الضوء الذي نسميه فوتونًا ، لذلك هذا عديم الكتلة ، وسوف يضرب الفوتون هذا الإلكترون ، وإذا كان للفوتون طاقة كافية ، يمكنه تحرير الإلكترون ، أليس كذلك؟ حتى نتمكن من التخلص منه ، ولذا اسمحوا لي أن أمضي قدما وأظهر ذلك هنا ، نظهر الإلكترون مفكوكًا وهكذا يتحرك الإلكترون ، دعنا نقول فقط ، هذا الاتجاه ، مع بعض السرعة ، v ، وإذا كان للإلكترون كتلة ، م ، نعلم أن هناك طاقة حركية. الطاقة الحركية للإلكترون سيساوي نصف mv تربيع. عادة ما يشار إلى هذا الإلكترون المحرّر الآن كضوئي. لذلك ينتج فوتون واحد ضوئيًا واحدًا. إذن يصطدم أحد الجسيمات بجسيم آخر. من حيث الفيزياء الكلاسيكية ، وإذا فكرت في ذلك يتم حفظ الطاقة. يمكنك التفكير فيه لذا فإن طاقة الفوتون والطاقة التي دخلت ، لذلك اسمحوا لي أن أكتب هذا هنا ، لذا فإن طاقة الفوتون ، الطاقة التي دخلت ، ماذا حدث لتلك الطاقة؟ اللازمة لتحرير الإلكترون. كان بعض من تلك الطاقة لذلك كان الإلكترون مرتبطًا ببعض الطاقة حرر الإلكترون. سأسمي هذا E لا شيء ، الطاقة التي حررت الإلكترون ، ثم باقي تلك الطاقة يجب أن يكون قد ذهب إلى الطاقة الحركية للإلكترون ، وحتى نتمكن من الكتابة هنا الكهروضوئية التي تم إنتاجها. الطاقة الحركية إذن ، الطاقة الحركية للإلكترون الضوئي. لنفترض أنك تريد حلها من هذا الإلكترون الضوئي. للطاقة الحركية لذلك سيكون هذا بسيطًا جدًا ، سيكون فقط الطاقة الحركية تساوي طاقة الفوتون طاقة الفوتون ، مطروحًا منه الطاقة اللازمة السطح المعدني. لتحرير الإلكترون من السطح المعدني وهذا لا شيء ، هنا أسميه Enaught ، قد تراها مكتوبة بشكل مختلف ، رمز مختلف لكن هذه هي وظيفة العمل. اسمحوا لي أن أكتب وظيفة العمل هنا ، ووظيفة العمل مختلفة لكل نوع من المعادن. إذن ، إنها أقل كمية من الطاقة هذا ضروري لتحرير الإلكترون ، ومن الواضح أن هذا سيكون مختلفًا اعتمادًا على المعدن الذي تتحدث عنه. حسنًا ، لنحل مشكلة.
الآن بعد أن فهمنا الفكرة العامة للتأثير الكهروضوئي ، دعونا نلقي نظرة على ما تطلبه منا هذه المشكلة. لذا فإن المشكلة تقول ، "إذا كان فوتون ذا طول موجي "525 نانومتر يضرب السيزيوم المعدني. وهذه هي دالة الشغل للسيزيوم المعدني. "ما هي سرعة إنتاج الفوتوإلكترون؟"لذا يريدون معرفة السرعة من الكهروضوئية المنتجة ، الذي نعلم أنه يختبئ في الطاقة الحركية هنا ، ونعرف أيضًا ما هي وظيفة الشغل. لذا فنحن نعرف ما هو E لا شيء هنا. ما لا نعرفه هو طاقة الفوتون لذلك هذا ما نحتاج إلى حسابه أولاً. وهكذا فإن طاقة الفوتون ، طاقة الفوتون تساوي ح ، وهو ثابت بلانك ، مرات التردد ، والتي عادة ما يرمز لها nu. أعطونا الطول الموجي لذا حصلنا على التردد ، لكن في المشكلة هنا. أعطونا الطول الموجي ، لذلك نحن بحاجة إلى الارتباط التردد إلى الطول الموجي ، وهذا مرتبط بـ c ، وهي سرعة الضوء ، يساوي لامدا ضرب نو. إذن ، c هي سرعة الضوء ، وهذا يساوي التردد ضرب الطول الموجي. لذا يمكننا التعويض بـ n عن التردد ، حسنًا ، لأننا نستخدم هذه المعادلة فقط ونقول أن التكرار متساوي إلى سرعة الضوء مقسومًا على الطول الموجي. التردد يساوي سرعة الضوء على لامدا ، حتى نتمكن من توصيل ذلك هنا ، والآن لدينا طاقة الفوتون يساوي hc على لامدا ، ويمكننا التعويض بهذه الأرقام. h هو ثابت بلانك ، وهو 6.626 ضرب 10 أس سالب 34. إذن ، ضرب 10 أس سالب 34 هنا. سرعة الضوء ، وهو 2.998 ضرب 10 أس 8 متر خلال ثوان ، وفي جميع أنحاء لامدا. لامدا هي الطول الموجي. هذا 525 نانومتر. إذن 525 ضرب 10 متر 9 سالب. حسنًا ، لنخرج الآلة الحاسبة للفوتون هنا. وحساب الطاقة لذا ، دعنا نمضي قدمًا ونجري هذه الحسابات ، لذلك لدينا 6.626 مرة 10 إلى سالب 34 وسنضرب هذا الرقم بسرعة الضوء 2.998 ضرب من 10 إلى 8 ونحصل على هذا الرقم. سنقسمه على الطول الموجي ، 525 مرة 10 إلى سالب 9 ونحصل على 3.78 مرة 10 أس سالب 19. لذا ، اسمحوا لي أن أمضي قدمًا وأكتب ذلك هنا. 3.78 في 10 أس سالب 19 وإذا كنت فعلت وحدات هنا ستحصل على الجول ، لذا دعونا نفكر في هذا الرقم لثانية ، 3.78 في 10 أس سالب 19 هي طاقة الفوتون. وطاقة الفوتون تلك أكبر من وظيفة العمل ، مما يعني إنه فوتون عالي الطاقة. إنه قادر على ضرب الإلكترون مجانًا ، لأن تذكر ، هذا الرقم هنا ، هو الحد الأدنى من الطاقة اللازمة لتحرير الإلكترون وهكذا تجاوزنا هذا الحد الأدنى من الطاقة ، ولذا فإننا ننتج ضوئيًا. إذن ، هذا الفوتون ذو طاقة عالية بما فيه الكفاية لإنتاج إلكترون ضوئي. لذلك دعونا نمضي قدمًا ونجد الطاقة الحركية من الإلكترون الضوئي الذي تم إنتاجه. إذن سنستخدم هذه المعادلة هنا. لذا دعني أذهب وأحصل على مساحة أكبر ، وسأعيد كتابة تلك المعادلة. إذن لدينا الطاقة الحركية للإلكترون الضوئي ، الطاقة الحركية للضوئية ، تساوي طاقة الفوتون ، طاقة الفوتون مطروحًا منها دالة الشغل. لذلك دعونا نعوض بالأرقام. كانت طاقة الفوتون 3.78 في 10 سالب 19 جول ثم وظيفة العمل هنا مرة أخرى ، إنه 3.43 ، لذلك ناقص 3.43 ضرب 10 أس سالب 19 جول. فلنخرج من الآلة الحاسبة مرة أخرى. إذن ، سنطرح من ذلك دالة الشغل 3.43 مرة 10 أس سالب 19 ونحصل على 3.5 ضرب 10 أس سالب 20. لذلك دعونا نمضي قدمًا ونكتب ذلك. هذا يساوي 3.5 ضرب 10 أس سالب 20 جول. هذا يساوي الطاقة الحركية من الفوتوإلكترون ، ونحن نعلم ذلك الطاقة الحركية تساوي نصف متر مربع. طلبت منا المشكلة حلها لسرعة الإلكترون الضوئي. لذا كل ما علينا فعله هو توصيل كتلة إلكترون وهو 9.11 مرة 10 إلى كيلوغرام 31 سالب ضرب تربيع. هذا يساوي 3.5 ضرب 10 أس سالب 20. لذا ، دعونا نجري هذه الحسابات. إذن نأخذ 3.5 في 10 أس سالب 20 نضرب ذلك في 2 ، ثم نقسم على كتلة الإلكترون ، 9.11 مرة 10 إلى 31 سالب وهذا يعطينا هذا الرقم ، الذي علينا أخذ الجذر التربيعي له. إذن ، الجذر التربيعي لإجابتنا يعطينا سرعة الإلكترون ، 2.8 ضرب 10 أس الخامس. لذلك إذا نظرت إلى المكان العشري هنا ، سيكون هذا واحد ، اثنان ، ثلاثة أربعة ، خمسة ، إذن 2.8 ضرب 10 أس الخامس متر في الثانية. إذن هذه هي سرعة إنتاج الإلكترون الضوئي ، 2.8 ضرب 10 أس الخامس متر في الثانية ، شدة هذا الضوء ، وإذا قمت بزيادة لذلك كان لديك المزيد من الفوتونات ، سوف ينتجون المزيد من الإلكترونات الضوئية. لذا يقوم فوتون واحد بإخراج إلكترون ضوئي واحد إذا كان لديه ما يكفي من الطاقة للقيام بذلك. لذلك دعونا نفكر في نفس المشكلة ، لكن دعونا نغير الطول الموجي. لذا ، ماذا لو تغير الطول الموجي الخاص بك إلى 625 نانومتر. إذن ماذا سيحدث الآن؟ لن أقوم بالحسابات ، حسنًا ، لتوفير الوقت ، أنا ولكن كل ما علينا القيام به يتم توصيل 625 هنا. لذا بدلاً من 525 ، عوض بـ 625 لحساب طاقتك ، وإذا فعلت ذلك ، لذلك إذا استخدمت 625 ضرب 10 أس سالب 9 هنا سأذهب وأعطيك الجواب فقط لتوفير بعض الوقت ، ستحصل على 3.2 مرة 10 أس سالب 19 جول. وهذا أقل من دالة الشغل. لذا اسمحوا لي أن أمضي قدمًا ، و 0 أبرز ذلك هنا. لذلك هذا الرقم ليس كذلك عالية مثل وظيفة العمل. كانت وظيفة العمل مقدار الطاقة كنا بحاجة لتحرير هذا الإلكترون ، وبما أن هذا أقل من دالة الشغل هذا يعني أننا لا نحصل على فوتوإلكترون. لذلك ، يجب أن يكون لديك فوتون ذو طاقة عالية بما يكفي من أجل إنتاج ضوئي. لا يهم حتى إذا قمنا بزيادة الكثافة. لذلك إذا كان لدينا المزيد والمزيد من هذه الفوتونات بهذا الطول الموجي ، أي الضوئية. ما زلنا لا ننتج وهكذا ، هذه هي فكرة التأثير الكهروضوئي ،و هو افضل شرح من خلال التفكير في الضوء كجسيم.
"الترجمه ترجمة قوقل" Good luck💪🏼
Nice
Neat
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Thank you. You are a kind person
I know this is a dumb question to ask, so please forgive my ignorance.....but how can a massless virtual particle (mediator of the Electromagnetic Force, for example) hit anything....let alone, a sub-atomic particle having mass....such as the Electron?
So a non-massive packet of energy can smash into something?....or is this just my misunderstanding?
Mass in itself is just how we perceive energy on a macroscopic level. Many RUclips channels cover this topic excellently, particularly PBS Space Time. Anyway, it's the energy of the photon that allows it to interact with the electron.
P.S, I'm no physicist so please take the time to check this out for yourself, I may be completely wrong :)
It's because the rest mass of photon is zero. But once it's moving it has a mass due to the kinetic energy (1/2)mv^2
There are two ways to think of it.
The first is that photons do have a certain kind of mass due to the equivalence of mass and energy, E=mc^2. Photons certainly carry energy, hence they carry mass. You're right that photons are massless with respect to INERTIAL mass. Inertial mass is basically the kind that prevents things from moving at the speed of light. The mass which is equivalent to energy is a bit different.
The second, and in some ways more accurate, way is to realize that in order for a photon to collide with and displace an electron, it doesn't need to have mass, it simply needs to impart momentum to the electron. But wait, isn't momentum determined from mass, p=mv? And if m=0 for photons then p=0 too? Not exactly. p=mv is not the right way to formulate momentum for quantum (small) objects like photons. Delve into the wonderful world of quantum mechanics and you'll find that momentum is described as an operator - basically it's a partial derivative - and its measurable, concrete values change dramatically based on the quantum system at hand. Sometimes it's equal to mv, but for other systems (like photons!), p=h/lambda.
The electron would already have a velocity before the photon hits it right? So the kinetic energy u calculated was actually not right!
In this case electrol was resting he got tired
Thanks Kahn Academy, you guys are so, so so helpful.
For more Khan Academy stuff, check out this: www.khanacademy.org
These videos are too quiet. I always have to turn up my volume for them
Türkçe altyazı yok mu
How does it depend on the intensity of light
The intensity of light only has to do with the photo current, since more photons are hitting the metal plate more electrons are ejected.
Great video!
good 👍👍👍👍👍👍👍👍👍👍
7:40 why didn't you half the mass of the electron because v=sqrt of ke*1/2 of mass
+(:Turretshooter500:) NO!, v= sqrt[(2 x KE)/mass] when he is rearranging the equation, you should move the 1/2 to the otehr side and divide whatever there is by it(the 1/2). Divinding by 1/2 is = to multiplying by 2 hence the 2 x KE on top
+Dennis Premoli Thanks, now I have a better understanding.
(:Turretshooter500:) NO probs, I have a test this friday so :)
Did you say “enon”?
Btw what programme do you use to make you're vids. Is the photoelectric effect anything related to E=MC^2 ??
(Well, quite obviously; I think I am being a little bit dumb)
+(:Turretshooter500:) nope.
+ronak dev Oh yeah, my mistake! Silly me📢📣📢
+(:Turretshooter500:) Idk we learned about E = MC^2 when we were learning about nuclear chemistry
+violinsheets do you think I care?
(:Turretshooter500:) Basically it says that mass and energy are interconvertible, usually applies to large losses of mass during nuclear decay. www.avogadro.co.uk/brief_topics/mass_energy.htm
Did he say photon is mass-less at 0.45? So he does not believe in E=mc^2 which says E has mass = E/c^2?
Besides its not a "right" kind of light with a "right" kind of frequency which knocks off an electron but ANY kind of light ABOVE a THRESHOLD frequency. One does not have to "tune" the frequency to knock out an electron. Of course the KE/electron will vary directly with supra threshold light freq
Photon is massless the energy and momentum of a photon depend only on its frequency (ν) or inversely, its wavelength (λ) and photon has momentum but it is massless because you can't use ordinary equation for calculating momentum for something that's moves the speed of light you have to use the calculations from special theory of relativity that said p = Planck's constant/ Lambda which is wavelength
your voice is really low
That E not is also called as threshold frequency
pavan bhallamudi E naught is the threshold ENERGY, not frequency. "nu naught" is threshold frequency
@Khan Academy, will this video be better if you increase the volume?
Thanks man.
how did the mass of the electron come up??
Why is the wavelength multiplied by 10^-9???
+Esperanza Vazquez Thats because is 525 nanometers, but since we are doing the calculation in meters per seconds...we need to convert nanometer into meter (10^9 nm = 1 m). If I ask you what is 1 Kg + 10 g, what would you say? 11 Kg or 11 g? You have to convert 1 Kg to 1000 g or vice versa
+Aryan Purohit are you teaching him yr.3-4 maths in an a level revision?
because you need to turn nanometar's in to metars
sir,can we see electron ?
No
Ironically, AAMC uses a calculator in its tutorial videos but denies its candidate the privilege of using the tool.
What do I do when they don't give me the work function in a question? I have a book that states them for all elements, but they give multiple values for differentent masses of the element, so I don't know which one to choose. I need the work function for sodium. Thanks for the video :)
work function=h*threshold frequency = h*c/wavelength threshold
I don't understand a lot of the terminology used in this video, any advice on videos that would explain them?, cause I'm lost
Patty Lilly you just need to really make an effort to conceptualize. I have a decent understanding of most of the theory just from reading wikipedia lol
Is this part of a high school or college program in the US?
dude its high school physics
In my physics class I learned it as E = hf, why so?
Should be the same thing, I learned it as E = Hv, i think the only difference between chem and physics is Hess's law?
@@mrfreak881 f and v are the same thing i.e frequency
we cannot see electron actually they are fixed in every atom. it is just sal's through which he is making electron
What happens to light(a photon) when it loses all its energy?
Thank you :)
Does photoelectric current depend on photo sensitivity of metal ?
How can it have energy without mass?
Use planck equation E=hf
Energy is not dependent on mass it depends on frequency
@@huzaifaabedeen7119 Theres a kg value in planks constant. frequency as in of atoms moving down a microscopic scale? you know what I'm saying rather than like a free falling object holding energy.
3.78-3.43 = 0.35 .(ten to minus19) not 3.5 to 10 to the minus 20 .........?
+John hart Isn't 0.35*10^-19 the same as 3.5*10^-20 which is in proper scientific notation?
+Benjamin Hsu of course.....thanks for your excellent videos
I had the same doubt for a while. But cleared now
Why does this occur rather than the electron becoming excited, moving up a shell, and then moving down, releasing a photon?