Hello there, professor Lewin! I'd like to say a big thank you: Your lectures, 801 and 802, were essential in my life to discover my love for physics, and my academical success is due largely to you! You helped me build a strong fundamental grasp of the concepts and they save me in uni to this day! Thank you so much for your effort, your attention and your lectures: part of me wouldn't be the same today if it weren't by them. Thank you so much!
Beste Walter. Ik zou u graag willen bedanken voor de manier waarop u fysica presenteert. Ik heb altijd geprobeerd mijn zoon te tonen wat de basis principles zijn van onze wereld zijn, gebaseerd op uw methodes. Hij is nu 16. De tijd vliegt. En hij is nu net door de tweede ronde van de Thaise fysica olympiade. U bent een inspiratie voor mij en vele anderen.
Answers: A: 22.4 MeV B: From Problem 215, the energy released in the fission of U235 is 173.2 MeV, which is about 7.7 times as much energy per reaction. However, compared to the fuel mass, this fusion reaction produces 2.8 Mev/amu, while the fission of U235 produces 0.737 MeV/amu. So while there's less energy for each reaction, the energy released per unit fuel mass is about 3.8 times as much for fusion as it is for fission. So this fusible fuel has more gravimetric energy density than fissible fuels like U235. C: This is not a possible reaction, because it requires an energy input, rather than produces an energy output. Energy addition would need to be given. This is consistent with what I expected prior to calculating, because fusion is typically the reaction for elements with atomic numbers less than that of iron. Supporting calculations: Approach: assume velocities are negligible, add up total mass before and after, and take difference. Convert from amu to kg (6.022*10^26 amu/kg), and then use E=m*c^2 to find corresponding energy. Then convert from Joules to MeV (1 J = 1.602e-13 MeV). Isotope mass data in amu: Lithium-6: 6.015122887 Hydrogen-2: 2.014101778 Helium-3: 3.016029322 Helium-4: 4.002603254 Berylium-7: 7.01692871 Mass balance: m_Li6 + m_H2 = 2*m_He + X Solve for X, and plug in data: X = m_Li6 + m_H2 - 2*m_He X = 0.024018157 amu X = 3.988*10^-29 kg E = X*c^2 E = 3.585*10^-12 J = 22.38 MeV For the Beryllium fission reaction to be possible, nucleons and charge must be conserved, and total mass must either stay the same or decrease. m_Be7 = m_He3 + m_He4 + Y Solve for Y: Y = m_Be7 - m_He3 - m_He4 Y = -0.001703866 amu, corresponding to -1.587 MeV The left-over mass (as energy) is negative, which implies it isn't a possible reaction.
a) Energy produced in the reaction is 23.8 MeV b) The energy liberated with uranium was -165 MeV. That means that in the reaction of lithium and deuterium gives a net energy positve that energy was gained, not released. c) The mass of borum is slightly less than the combined mass of the products, meaning this reaction is not energetically favorable under normal conditions.
A)22.36MeV B)I'm ashamed I wasn't able to do it as it will be taught to us after JEE Mains C)The Qvalue came out to be -1.527MeV suggesting it is highly Endothermic rxn, so no, it will not take place.
a) 22,373 MeV, just the binding energy delta b) Per Atom it is significantly less but per nucleon a lot more. c) It results in positive binding energy (13.85 MeV ) but the energy required to break the nucleus apart is higher. I believe this reaction can happen, however, since there is no neutron absorption and Be4 is stable, a violent event is needed to break the nucleus apart. For example a high energy collision with another Be4. Big bangs, stars exploding ... maybe in a particle collider. Outside such conditions I believe it will not happen.
a. 22.4 MeV. b. The energy released per nucleon (or per grams) in Li6-H2 fusion is 3.8 times higher than that in U235 fission. c. Yes, but *not spontaneously*. The reaction requires extra energy input to occur (ΔE = -1.6 Mev), making it energetically unfavorable.
a. 22.5 MeV is released in the fusion reaction of Li(6) + H(2) -> 2He(4) b. ~8-9 times less than the fission of U(235) for 1 atom. But there are 235/8 more atoms in the Fusion reaction of the same mass of reactants. Hence it will about 30 times 22.5. So overall the Fusion reaction is about 3-4 times more energy. c. The binding energy is almost neck to neck in Be(7) -> He(4) + He(3), but slightly higher on the left by ~ 1Mev. Therefore the reaction is not favored.
For this problem I googled binding energies of the various components - easier than calculating from masses as I did in previous problems. Part (a) Binding energy Of Li6 = 31.99 MeV, H2 = 2.23 MeV; total = 34.22 MeV. Binding energy of (2) He4 = 2x28.296 Me = 56.59 Mev Net binding energy increase = 22.37 MeV, or 2.80 MeV per nucleon. Part (b) compare to fission of U235 From previous problem, net energy was 173.2 MeV (including product neutrons kinetic energy), but note net energy per nucleon was only 0.72 MeV. The fusion reaction is 3.83 times stronger per unit mass. Part (c) This reaction will not occur spontaneously because the final binding energy is less than the initial binding energy. Be7 -> 37.6 MeV Products He4 => 28.3 MeV, He3 -> 8.48 MeV, total 36.78 MeV Thus, this reaction requires an input of 0.82 MeV to proceed.
22.17MeV? An order of magnitude lower than fission reactions of U235. As for (C), I originally thought it could take place, but looking at this fresh again, I think I was mistaken. The total mass of the product (He-4 and He-3) is 7.0186 u, and the mass of Beryllium-7 is 7.0169 u. So the product is 0.0017 u greater, meaning it would need additional energy to proceed. Posting this somewhat hesitantly given the threat of removing such questions in future, and not being totally certain of my answers.
you should not compare 1 U235 atom with 1 nucleus of Li6 and H2. You should compare the same mass - thus e.g., 1 gram of U235 with 1 gram of the Li6 and H2
a) 22.1 MeV per nucleus b) Energy per gm from U235 is 4 times less than energy per gm of Li6 c)Binding energy per nucleon of He3 is less than the binding energy of Be7 making He3 unstable. So it won't take place I'm not sure about the answer.
Even if the next problem will be "boring newtonian mechanics": With Prof.Lewin it is NEVER boring! Remember the words of him: Teaching boring physics is a crime.
I will never be boring. Newtonian Mechanics Problems don't have to be boring either. *I should have left the word "boring" out.* The fact remains that so far there are only 5 correct solutions to problem 216. *If there will be fewer than 12 then I will stop Nuclear Physics Problems.*
a) 22.4 MeV b) Fission of U235 released approximately 7.5 times the energy released by the fusion of Li6 and H2 c) No, because the binding energy per nucleon of He3 (2.6 MeV) is less than the binding energy per nucleon of Be7 (5.4 MeV), thus making the He3 unstable. So the fission will not take place. Sir please keep the nuclear physics questions continued, they're really fun to do.
comparing fission of 1 U235 nuleus with fusion of 1 Li6 + 1 H2 nucleus is not the right thing to do. You have to compare 1 gram of U235 with 1 gram of Li6+H2. In other words you have to compare the same number of nucleons on both sides.
@@lecturesbywalterlewin.they9259 Since we are trying to get same number of nucleons, I think it would be the same if I get the energies in MeV/amu to avoid getting the Avagadro's number and all that involved (my chemistry is not good). In case of uranium, since one atom has mass of 235 amu and releases energy of 168 MeV, the energy per amu would be 168/235. In case of Li6 + H2, I took the sum of their masses U235: 0.8 MeV/amu Li6 + H2: 2.8 MeV/amu So the ratio of (Li6 + H2)/U235 is 3.5 This was quite surprising to me, because I thought uranium would've had more than Li6 and H2. I'll try it again to confirm what I did is correct, by taking it in grams. Please tell if my logic is correct and the answer too.
@@lecturesbywalterlewin.they9259 Second attempt I tried getting them in grams. I'll try to explain my logic by taking the case of uranium. I multiplied the energies by 6.022 * 10^23 to get energy released by one mole or by 235 g of uranium. Then I divided by 235 to get the energy released by 1 gram. And so here are the answers: U235: 4.3 * 10^23 MeV/g Li6 + h: 16.7 * 10^23 MeV/g Edit: I realized that I just did the unit conversions the long way. I'm such an idiot.
@@lecturesbywalterlewin.they9259 Sir, just wanted to ask if you could see my comment? Because it isn't showing up when I see from other accounts. Same thing happened with my other comments but they showed up when you replied. So just wanted to know.
Greetings Professor! I would like to know which edition of Giancoli was used in the 8.02 lectures because I can't find the diagrams in the assignment questions.
I will give it a try: a) 31.9965+2.2245 MeV = 56.594 MeV -22.37MeV. So per 6Li and 2H fusion, 22.4MeV are released. b) For 235U we had 165MeV per fission. Let's compare that per mass: 165MeV/235.044dalton * x = 22.37MeV/(( 6.02+2.01)dalton) hence x= 4.0. So per gram, the fusion releases 4x more energy than 235U. Even more scary or useful. Depending on humans... c) This fission doesn't happen. 37.600Mev = 7.7186MeV + 28.297MeV+ deltaE= 36.02MeV + 1.58MeV. This fission would absorb delta E= 1.58MeV per 7Be nucleus. I would rather expect, that the two He nuclei on the right hand side would fuse to 7Be. Then, 1.6MeV per fusion would be released. So, as written, the reaction doesn't happen. I hope, you nevertheless continue. Even if my answer is wrong and not contributing to 12 correct answers.
Dear Prof.Lewin, I might have an idea for a future problem. Maybe you read it and think it's rubbish, but here we go: We know that the binding energy per nucleon has a maximum at iron. The first element existing in the universe was H. Through fusion in stars, He, Li,..., up to Fe have formed. But how can we explain that heavier nuclei than Fe have formed? The binding energy per nucleon decreases at elements heavier than iron. So: Where did all the Pb, U, Au, Pt etc. come from?
the answer to this problem is too easy. High Z elements have all been formed in Supernova explosions. All the carbon in your body came from Supernovae - use gogle
@@lecturesbywalterlewin.they9259 Yes, I know... maybe others didn't? But how does that happen in Supernovae? Fusions to above Fe don't "just happen" like the He to O cycle in stars.
a) 22MeV per nucleus b) The energy per gr from Li fusion is 4 times grater than the energy from U235 fission. c) Yes. The binding energy of Be7 nucleus is 37.6 while the binding energy of He4 + He3 is 36MeV. Well, I am not sure.
Professor, I'm Sourav from India. I'm a physics graduate. 26yrs old. I'm working as a private tuition teacher of high school standard. I want to demonstrate all the experiments to my students. But there has two problems, one is cost of the experimental setup; second one is experimental knowledge in all chapters. I want a suggestions from you, if you know any book which can help me gain knowledge as well as demonstrate physics experiment using home's setup spending very few money.... Plz let me know.... If you allow me any other form of direct contact with you!,i will let you know what type of experiment I'm talking about & related chapter.... So that you can guide me.... Desperately i need this help. Touching your feet.... Have a good time professor....
For those who think that nuclear energy is safe... buy a house in Chernobyl or Fukushima and try to be happy living there. Or be a good human being and live in Hiroshima or Nagasaki. And contemplate the maximum that human intelligence can offer after 80 years. Remember, GOD created everything to keep us safe. Unfortunately, we haven't managed to respect the planet or live together in harmony to perpetuate... Amem.
If I do not get at least 12 correct answers on Problem 216 then I will stop nuclear physics problems. So far there ar 5 viewers who have all 3 questions correct. I have NOT posted those 5 solutions yet. I will do that on the day that I post the solutions
Hello there, professor Lewin! I'd like to say a big thank you: Your lectures, 801 and 802, were essential in my life to discover my love for physics, and my academical success is due largely to you! You helped me build a strong fundamental grasp of the concepts and they save me in uni to this day! Thank you so much for your effort, your attention and your lectures: part of me wouldn't be the same today if it weren't by them. Thank you so much!
You are most welcome.
Beste Walter. Ik zou u graag willen bedanken voor de manier waarop u fysica presenteert. Ik heb altijd geprobeerd mijn zoon te tonen wat de basis principles zijn van onze wereld zijn, gebaseerd op uw methodes. Hij is nu 16. De tijd vliegt. En hij is nu net door de tweede ronde van de Thaise fysica olympiade. U bent een inspiratie voor mij en vele anderen.
dat is heel fantastisch!!!
Happy to see you again sir
God Bless you. You are the guy who proved to me that I can Love physics! Don't know how to thank you professor!❤❤❤❤❤❤
It's my pleasure
Love from India ❤❤❤❤
Am an arts student but love physics and love your content
Plz never stop posting
Answers:
A: 22.4 MeV
B: From Problem 215, the energy released in the fission of U235 is 173.2 MeV, which is about 7.7 times as much energy per reaction. However, compared to the fuel mass, this fusion reaction produces 2.8 Mev/amu, while the fission of U235 produces 0.737 MeV/amu. So while there's less energy for each reaction, the energy released per unit fuel mass is about 3.8 times as much for fusion as it is for fission. So this fusible fuel has more gravimetric energy density than fissible fuels like U235.
C: This is not a possible reaction, because it requires an energy input, rather than produces an energy output. Energy addition would need to be given. This is consistent with what I expected prior to calculating, because fusion is typically the reaction for elements with atomic numbers less than that of iron.
Supporting calculations:
Approach: assume velocities are negligible, add up total mass before and after, and take difference. Convert from amu to kg (6.022*10^26 amu/kg), and then use E=m*c^2 to find corresponding energy. Then convert from Joules to MeV (1 J = 1.602e-13 MeV).
Isotope mass data in amu:
Lithium-6: 6.015122887
Hydrogen-2: 2.014101778
Helium-3: 3.016029322
Helium-4: 4.002603254
Berylium-7: 7.01692871
Mass balance:
m_Li6 + m_H2 = 2*m_He + X
Solve for X, and plug in data:
X = m_Li6 + m_H2 - 2*m_He
X = 0.024018157 amu
X = 3.988*10^-29 kg
E = X*c^2
E = 3.585*10^-12 J = 22.38 MeV
For the Beryllium fission reaction to be possible, nucleons and charge must be conserved, and total mass must either stay the same or decrease.
m_Be7 = m_He3 + m_He4 + Y
Solve for Y:
Y = m_Be7 - m_He3 - m_He4
Y = -0.001703866 amu, corresponding to -1.587 MeV
The left-over mass (as energy) is negative, which implies it isn't a possible reaction.
a) Energy produced in the reaction is 23.8 MeV
b) The energy liberated with uranium was -165 MeV. That means that in the reaction of lithium and deuterium gives a net energy positve that energy was gained, not released.
c) The mass of borum is slightly less than the combined mass of the products, meaning this reaction is not energetically favorable under normal conditions.
If Earth wasn't geocentric, stationary and immovable with electrostatics at it's core then none of this would even be plausible.
Thank you, Professor.
You are welcome.
A)22.36MeV
B)I'm ashamed I wasn't able to do it as it will be taught to us after JEE Mains
C)The Qvalue came out to be -1.527MeV suggesting it is highly Endothermic rxn, so no, it will not take place.
a) 22,373 MeV, just the binding energy delta b) Per Atom it is significantly less but per nucleon a lot more. c) It results in positive binding energy (13.85 MeV ) but the energy required to break the nucleus apart is higher. I believe this reaction can happen, however, since there is no neutron absorption and Be4 is stable, a violent event is needed to break the nucleus apart. For example a high energy collision with another Be4. Big bangs, stars exploding ... maybe in a particle collider. Outside such conditions I believe it will not happen.
You're reminding me one ukrainian physics teacher
the energy liberated in the fusion reaction is approximately 17.6 MeV.
what are yoour asnwers to b) and c)?
Hiii sir ❤❤❤ love from Nepal 🇳🇵💖💖😊
One of the cons of the Medical school is stop watching you as I used to. miss you prof from 🇪🇬
do what is best for you
WALTER LEWIN SIR ❤
a. 22.4 MeV.
b. The energy released per nucleon (or per grams) in Li6-H2 fusion is 3.8 times higher than that in U235 fission.
c. Yes, but *not spontaneously*. The reaction requires extra energy input to occur (ΔE = -1.6 Mev), making it energetically unfavorable.
Great Sir ❤
a. 22.5 MeV is released in the fusion reaction of Li(6) + H(2) -> 2He(4)
b. ~8-9 times less than the fission of U(235) for 1 atom. But there are 235/8 more atoms in the Fusion reaction of the same mass of reactants. Hence it will about 30 times 22.5. So overall the Fusion reaction is about 3-4 times more energy.
c. The binding energy is almost neck to neck in Be(7) -> He(4) + He(3), but slightly higher on the left by ~ 1Mev. Therefore the reaction is not favored.
what counts is NOT 1 U 235 vs 1 Li and 1 H2 - what counts is the same mass - thus e.g., 1 gram vs 1 gram
For this problem I googled binding energies of the various components - easier than calculating from masses as I did in previous problems.
Part (a)
Binding energy Of Li6 = 31.99 MeV, H2 = 2.23 MeV; total = 34.22 MeV.
Binding energy of (2) He4 = 2x28.296 Me = 56.59 Mev
Net binding energy increase = 22.37 MeV, or 2.80 MeV per nucleon.
Part (b) compare to fission of U235
From previous problem, net energy was 173.2 MeV (including product neutrons kinetic energy), but note net energy per nucleon was only 0.72 MeV. The fusion reaction is 3.83 times stronger per unit mass.
Part (c) This reaction will not occur spontaneously because the final binding energy is less than the initial binding energy.
Be7 -> 37.6 MeV
Products He4 => 28.3 MeV, He3 -> 8.48 MeV, total 36.78 MeV
Thus, this reaction requires an input of 0.82 MeV to proceed.
Hi bro. Can you say me something? How did you send the solution to sir?
22.17MeV? An order of magnitude lower than fission reactions of U235. As for (C), I originally thought it could take place, but looking at this fresh again, I think I was mistaken. The total mass of the product (He-4 and He-3) is 7.0186 u, and the mass of Beryllium-7 is 7.0169 u. So the product is 0.0017 u greater, meaning it would need additional energy to proceed. Posting this somewhat hesitantly given the threat of removing such questions in future, and not being totally certain of my answers.
you should not compare 1 U235 atom with 1 nucleus of Li6 and H2. You should compare the same mass - thus e.g., 1 gram of U235 with 1 gram of the Li6 and H2
a) 22.1 MeV per nucleus
b) Energy per gm from U235 is 4 times less than energy per gm of Li6
c)Binding energy per nucleon of He3 is less than the binding energy of Be7 making He3 unstable. So it won't take place
I'm not sure about the answer.
Even if the next problem will be "boring newtonian mechanics": With Prof.Lewin it is NEVER boring! Remember the words of him: Teaching boring physics is a crime.
I will never be boring. Newtonian Mechanics Problems don't have to be boring either. *I should have left the word "boring" out.* The fact remains that so far there are only 5 correct solutions to problem 216. *If there will be fewer than 12 then I will stop Nuclear Physics Problems.*
Love ypu sir from imdia
Bro atleast edit your spelling mistakes
Bro correct it.
Thanks Professor Lewin for All Your Inspiring Work!!! 👍😉🇳🇱
a) 22.4 MeV
b) Fission of U235 released approximately 7.5 times the energy released by the fusion of Li6 and H2
c) No, because the binding energy per nucleon of He3 (2.6 MeV) is less than the binding energy per nucleon of Be7 (5.4 MeV), thus making the He3 unstable. So the fission will not take place.
Sir please keep the nuclear physics questions continued, they're really fun to do.
comparing fission of 1 U235 nuleus with fusion of 1 Li6 + 1 H2 nucleus is not the right thing to do. You have to compare 1 gram of U235 with 1 gram of Li6+H2. In other words you have to compare the same number of nucleons on both sides.
@@lecturesbywalterlewin.they9259 Since we are trying to get same number of nucleons, I think it would be the same if I get the energies in MeV/amu to avoid getting the Avagadro's number and all that involved (my chemistry is not good). In case of uranium, since one atom has mass of 235 amu and releases energy of 168 MeV, the energy per amu would be 168/235. In case of Li6 + H2, I took the sum of their masses
U235: 0.8 MeV/amu
Li6 + H2: 2.8 MeV/amu
So the ratio of (Li6 + H2)/U235 is 3.5
This was quite surprising to me, because I thought uranium would've had more than Li6 and H2. I'll try it again to confirm what I did is correct, by taking it in grams.
Please tell if my logic is correct and the answer too.
@@lecturesbywalterlewin.they9259 Second attempt
I tried getting them in grams.
I'll try to explain my logic by taking the case of uranium. I multiplied the energies by 6.022 * 10^23 to get energy released by one mole or by 235 g of uranium. Then I divided by 235 to get the energy released by 1 gram.
And so here are the answers:
U235: 4.3 * 10^23 MeV/g
Li6 + h: 16.7 * 10^23 MeV/g
Edit: I realized that I just did the unit conversions the long way. I'm such an idiot.
@@lecturesbywalterlewin.they9259 Sir, just wanted to ask if you could see my comment? Because it isn't showing up when I see from other accounts. Same thing happened with my other comments but they showed up when you replied. So just wanted to know.
@@Tokito62257 Bro please answer me. Have you only posted the solution in the comment section or also sent to sir's home address?
Greetings Professor! I would like to know which edition of Giancoli was used in the 8.02 lectures because I can't find the diagrams in the assignment questions.
8.02
Physics for Scientists & Engineers by Douglas C. Giancoli.
Prentice Hall
Third Edition
ISBN 0-13-021517-18
@@lecturesbywalterlewin.they9259 Thank you professor.
Love from Bangladesh sir🎉
Sir happy post diwali 🎇
As i think you were very busy for good questions as you didn't make a video on diwali
Btw love from india ❤
I will give it a try:
a) 31.9965+2.2245 MeV = 56.594 MeV -22.37MeV.
So per 6Li and 2H fusion, 22.4MeV are released.
b) For 235U we had 165MeV per fission. Let's compare that per mass: 165MeV/235.044dalton * x = 22.37MeV/(( 6.02+2.01)dalton) hence x= 4.0. So per gram, the fusion releases 4x more energy than 235U. Even more scary or useful. Depending on humans...
c) This fission doesn't happen.
37.600Mev = 7.7186MeV + 28.297MeV+ deltaE= 36.02MeV + 1.58MeV. This fission would absorb delta E= 1.58MeV per 7Be nucleus. I would rather expect, that the two He nuclei on the right hand side would fuse to 7Be. Then, 1.6MeV per fusion would be released. So, as written, the reaction doesn't happen.
I hope, you nevertheless continue. Even if my answer is wrong and not contributing to 12 correct answers.
Dear Prof.Lewin, I might have an idea for a future problem. Maybe you read it and think it's rubbish, but here we go:
We know that the binding energy per nucleon has a maximum at iron. The first element existing in the universe was H. Through fusion in stars, He, Li,..., up to Fe have formed.
But how can we explain that heavier nuclei than Fe have formed?
The binding energy per nucleon decreases at elements heavier than iron. So: Where did all the Pb, U, Au, Pt etc. come from?
the answer to this problem is too easy. High Z elements have all been formed in Supernova explosions. All the carbon in your body came from Supernovae - use gogle
@@lecturesbywalterlewin.they9259 Yes, I know... maybe others didn't?
But how does that happen in Supernovae? Fusions to above Fe don't "just happen" like the He to O cycle in stars.
a) 22MeV per nucleus
b) The energy per gr from Li fusion is 4 times grater than the energy from U235 fission.
c) Yes. The binding energy of Be7 nucleus is 37.6 while the binding energy of He4 + He3 is 36MeV. Well, I am not sure.
c) THINKKKKKK *the BE went down*
@@lecturesbywalterlewin.they9259 Wll, now I am sure. The answer to c) question is NO, because energy is getting down.
thank you professor
You are very welcome
SIR I want to hug you you are best sir
Thanks sir
Sir i weak in Newtonian Mechanics...So,What should i do??
eat yogurt every day but *never on Fridays* that also worked for Einstein and for me
@lecturesbywalterlewin.they9259 Sir, if I do this, I can do physics well?
Sir i am glade to see your video
Hope you see my comment
Iam in 9 grade but love your vidoe I hope i will meet you😊
Cannot find the binding energy for Be7
use google
@@lecturesbywalterlewin.they9259 I have done that. I used even Gemini.
@@YannisAlepidis Change your search words a little. It really is easy to find.
Search Chemlin Be7
Hello sir 🎉🎉
Sir please help me out on that potential energy problem?!
watch my 8.02 lectures. That's what they are for - your answers are all there.
Professor, I'm Sourav from India. I'm a physics graduate. 26yrs old. I'm working as a private tuition teacher of high school standard. I want to demonstrate all the experiments to my students. But there has two problems, one is cost of the experimental setup; second one is experimental knowledge in all chapters. I want a suggestions from you, if you know any book which can help me gain knowledge as well as demonstrate physics experiment using home's setup spending very few money.... Plz let me know....
If you allow me any other form of direct contact with you!,i will let you know what type of experiment I'm talking about & related chapter.... So that you can guide me....
Desperately i need this help.
Touching your feet.... Have a good time professor....
you can only contact me here in my chanel.
For those who think that nuclear energy is safe... buy a house in Chernobyl or Fukushima and try to be happy living there.
Or be a good human being and live in Hiroshima or Nagasaki. And contemplate the maximum that human intelligence can offer after 80 years.
Remember, GOD created everything to keep us safe.
Unfortunately, we haven't managed to respect the planet or live together in harmony to perpetuate...
Amem.
sir please dont stop nuclear physics first one is approx 22.3K MeV
If I do not get at least 12 correct answers on Problem 216 then I will stop nuclear physics problems. So far there ar 5 viewers who have all 3 questions correct. I have NOT posted those 5 solutions yet. I will do that on the day that I post the solutions
❤❤
How old are you honorable teacher
stone age
I LOVE YOU SIR YOU ARE MY IDLE
I am from india
Come to India and join iit bombay
U mean as a professor ryt?
@@devnallacherry08 Ya.
Walter, why do you love physics? why should one study physics?
I was born for Physics and Physics was made for me
❤❤❤❤❤❤❤❤❤❤❤
:)
First!
Sir plz dubbed hindi
dubbing 1 50 min lecture takes about 10 hours - If you dub them, please send your dubbed version to me and I will upload them
Same love from INDIA
PREPARING FOR IIT JEE 2026
good luck!