Atmospheric Ignition & The Fate of the World | Summer School '26 | Horizon IIT Madras

[00:09:25]World War II was the most devastating conflict in human history with a death toll of over 70 million people. The war finally came to an end with the bombing of Hiroshima and Nagasaki.

[00:09:47]In the second year of the war, the Manhattan project was started to produce nuclear weapons in hopes of ending the war sooner. Their idea was to create an atomic bomb that would split heavy uranium 235 atoms to release energy.

[00:10:12]Did you know that the peak temperature that was observed at the nuclear vision explosion of Hiroshima bomb was approximately 60 million° C.

[00:10:27]To put this into perspective, the temperature of our sun's center is 15 million° C.

[00:10:37]That means this temperature was almost four times larger.

[00:10:52]This posed a serious question. Physicist Edward Teller asked, "If we create temperatures hotter than the center of the sun in our atmosphere, will that heat ignite the air itself?

[00:11:10]This um possibility terrified scientists and a committee of three scientists were given the task of verifying this C. Marvin Edward Teller and Emil Konopinsky.

[00:11:29]They published the paper LA62, ignition of the atmosphere with nuclear bombs, detailing this possibility and how they disproved it before the Trinity test.

[00:11:42]Today, more than 60 years later, LA 692 remains a beautiful example of scientific uh reasoning under extraordinary certainty. And in today's summer school session, we will retrace their journey to understand how they proved that the atmosphere will not ignite.

[00:12:04]Now, Sanankit will take over and give us a brief overview.

[00:12:11]Good evening all. So, from what Olev has explained so far, I guess you guys have an idea of the gravity of the situation they were in. The Manhattan project was underway. It was weeks away from the Trinity test. For your information, the Trinity test was the first successful nuclear explosion in human history. And then someone asks an important question.

[00:12:34]The question was, what if the bomb not just destroy a city, not just kill thousands, but set the entire atmosphere on fire? A chain reaction propagating from the bomb outward until there is no atmosphere left. The question was so serious enough that a paper on this they arrived at a conclusion. Today's session is going to be about that. Now before we begin we need to speak this CP we need to speak the language of its paper so that you guys have a better understanding. You guys have to keep in mind that this paper was written by three of the best physicists in the 20th century and hence they did not stop to explain the terms they used nor did they motivate any equations that was used.

[00:13:33]So before we begin I would like to make you guys familiarize with a few concepts that will come handy in the later parts of the session.

[00:13:44]Okay, imagine that you're in a completely dark room. You cannot see anything and you're throwing a tennis ball randomly in all direction. Assume that somewhere in the room there is an object and you want to know how likely you are to hit it with a randomly thrown ball. What is the probability that you strike it successfully?

[00:14:05]Obviously, the answer depends on how big the object is. Yes, the size of the object, how much area it presents to you when you throw the ball. If it is a large flat wall, it is easy to hit. If it's a vertical pole, it's harder. If it's a single point, it's nearly impossible to strike that object.

[00:14:26]Now, this idea can be brought into nuclear physics as well. The probability of hitting the object depends on the area it presents. When it comes to nuclear physics, the object is not a physical barrier. Rather, it is an interaction. What if instead of hitting a wall, you want to know how likely a moving particle is to interact with a target particle in a specific way. Maybe it will get scattered off or maybe it will be absorbed by it or a chemical reaction can happen. This probability can actually be calculated by assigning an effective area to each target particle.

[00:15:05]If the interaction is very likely to occur, the cross-section is huge. If it is less, well, the cross-section will be smaller as well. Now, you would have come across the idea of area of cross-section in other parts of physics such as electromagnetism when you were dealing with flux. This idea of cross-section in nuclear physics is quite different from it because here it need not always be equal to the geometric area.

[00:15:32]If you look at the slide given here, in the first situation, the effective area is actually equal to the geometric cross-section that is p<unk> r². But in the other two cases, it may or may not be equal to that. It can be greater as well as less than that. In fact, the cross-section is not really a property of the particle alone. It depends on the chemical reaction, depends on how much energy the particles have.

[00:15:59]The cross-section is measured in a unit called barn where one barn is equal to 10 ^ -24 cm squared. Now this idea will be used throughout the session. So keep this in mind.

[00:16:15]The next thing is something called bmstral.

[00:16:19]Perhaps you would be familiar with this from your high school physics. We all know that an accelerating charge emits electromagnetic radiation because the accelerating charge disturbs the electromagnetic field around it. This disturbance propagates outward as a wave as a radiation. Now suppose you have a very hot volume a very hot volume of gas. The gas contains a large number of electrons and nuclei. Now the temperature is large indicates that the electrons are moving at enormous speeds at speeds that are significant fractions of the speed of light. As they are moving through the gas they are constantly deflected by the electric field of the surrounding nuclei. So this deflection causes them to decelerate. As they decelerate their speed decreases, their direction changes. This decrease in kinetic energy is actually released as electromagnetic radiation.

[00:17:12]Hence we coined the term broomstral lang for this phenomena that is em radiation emitted when a fastm moving charged particle such as an electron rapidly decelerates or is deflected by another charged particle. This is a very crucial concept in this in today's discussion because this is one of the major ways in which energy is lost after a nuclear reaction happens. A detailed discussion upon this will be done in the later parts of the session. So keep this in mind.

[00:17:52]Now a quick briefing of what today's discussion is going to be. First we'll be discussing about the energy production during the bomb blast.

[00:18:02]When nitrogen nuclei in the air react with each other at extreme temperatures they release energy. Well, how much energy do they release? How fast does it happen? Under what conditions? And why exactly did we choose nitrogen nuclei as the most probable candidate for this reaction? All of this will be discussed in next in this part. In the second part, we'll be talking about the energy losses involved.

[00:18:26]Well, along with the production of reaction, production of energy, the hot electrons in air are radiating energy away at an enormous rate. So can the production ever keep up with the loss?

[00:18:38]And is there any temperature at the which at which the reaction actually becomes self sustaining?

[00:18:45]And the third part is about a theoretical possibility.

[00:18:50]Setting aside all the physics involved, can any real bomb supply enough energy to heat enough air to even attempt ignition?

[00:19:00]So these three questions cover the entire architecture of today's session.

[00:19:05]Now Satik will be taking over the session.

[00:19:13]Yeah. So for any scenario like for any kind of good scenario or a bad scenario be it be helpful to humans or harmful to humans they all the scientists always try to find out the ideal case because ideality in reality does not exist. But if we get the overview of ideal case, we can estimate the real chances of it happening. Similarly, in our atmospheric ignition case, we estimate the best ideal case, the best as in the very absurd estimations, very high values of energies and on the whole the worst case, the best worst case scenario.

[00:20:03]Keep this in mind because this is the sole thing sole basis of the entire paper. The best worst case scenario.

[00:20:14]Imagine you own a business. Now, in order to run your business smoothly and earn profits no matter what, you need data. Data of your products being sold, the cost, the production cost, the managerial cost and all. Then you would need to assume some conditions. The worst possible conditions may be some stock market crashed or maybe some new virus and pandemic has happened.

[00:20:43]Whatever in even in such scenarios your company should be able to thrive at least able to maintain not go into losses. In that way you will be able to run your business smoothly. Similarly, the atmospheric ignition scenario is no different than this business scenario.

[00:21:03]Essentially, the data we collect here is related to atoms, their energies and the cascading effects and we assume the worst possible condition.

[00:21:16]Now, in order to understand how we proceed, let us consider a system. A system it can be an engine or a turbine or anything. It consumes some form of energy and it produces another form of energy.

[00:21:36]Now if the consumed form of energy is greater than what is being produced the system cannot sustain itself.

[00:21:46]If the produced energy exceeds the consumed form of energy then the system essentially use the produced energy to feed itself on and sustain our atmosphere in this case is the system. So basically the energy produced in the subsequent fusion reactions should be able to sustain it and effectively release enormous amounts of energy and transferring the energy throughout the atmosphere thereby causing the atmospheric ignition.

[00:22:21]Now at for the atmospheric ignition scenario we'll we have to consider two things energy suppliers and energy consumers. Now energy is primarily supplied by the atomic bomb. The then purpose of being built the atomic bomb and due to its immense power and immense nuclear energy it produces vast amounts of energy in within very few fractions of seconds thereby raising the temperature to very high values.

[00:23:02]Now this report talks about in such scenarios where the temperature has risen such dramatically could the could the conditions be satisfied for subsequent fusion reactions and thereby igniting the atmosphere.

[00:23:22]From our lower classes we have learned or memorized this thing about the composition of earth's atmosphere.

[00:23:31]Earth's atmosphere primarily consists of nitrogen with 78%, oxygen 21%, argon.9% and remaining gases.

[00:23:42]Now in order for our best worst case scenario, we consider only lighter elements. Lighter elements such as oxygen, nitrogen, carbon and all. Why?

[00:23:51]because their fusion their nuclei provide us the possibility of nuclear fusion releasing tremendous amounts of energy.

[00:24:03]Now out of those lighter elements we have to choose particularly the worst element. Now the worst element which we chose for our best worst case scenario is nitrogen. Why? Because it is unstable compared to the rest of the nuclear.

[00:24:19]Oxygen is very much stable and even if we choose oxygen its fusion reactions do not provide that much of an energy when compared to the nitrogen's fusion reactions. You can see two reactions being marked with arrows. One is a radiative capture other is a alpha emission.

[00:24:39]Keep these reactions in mind as we'll discuss them in later slide and why these reactions are important.

[00:24:47]Essentially the five types of reactions which can be possible were mentioned in the paper LA 62 which are being showcased in the slide.

[00:25:03]Now for fusion reaction to takes place the atoms or the nitrogen nuclear need to come closer. But you from your + 1 +2 education you might have learned about electrostatics like charges repel. Similarly each nitrogen atom consists of an electron cloud. So when they try to come closer these electron clouds try to repel each other creating a some sort of columic repulsions and a barrier between them. Now the fusion can take place only after crossing this barrier or this fence gate. Now this barrier is also known as kulum barrier which is about 9.6 mega electron volts. Now 9.6 mega electron volts seems like a harmless number you think but in order to understand that consider the room in which you are listening to this session. Its ambient temperature is about 300 kel. To put that into perspective, it it amounts to an energy of 0.025 electron volts.

[00:26:16]Still not satisfied with this explanation? Well, take sun score whose temperature is about 1.5 into 10 ^ 7 kel. Now, that temperature corresponds to an energy of 0.13 mega electron volt.

[00:26:33]So even the sun's core temperature is lower comp that temperature which corresponds to such energies is lower compared to that of coolum barrier.

[00:26:45]The conversion factor I have written it down in the slide where one electron voltage is equal to is implies 11,65 kel.

[00:27:02]Now we have two possibilities.

[00:27:04]The colliding nuclei have energy greater than coolum barrier. The colliding nuclei have less than the coolum barrier energy. But you will ask what is the energy of colliding nuclei? What are we considering the energy of particle A the energy of particle B or both?

[00:27:30]Well, for a simplistic view, we are considering two particles A and B moving with velocities VA and VB. These velocities are less than the speed of light.

[00:27:45]Although in our case they are very close to speed of light for simplicity and explanation sake.

[00:27:52]Now from a classical mechanics or Newton Newtonian mechanics you must have studied about systems of particles and where for a two particle system kinetic energy can be written as kinetic energy of of center of mass plus kinetic energy with respect to center of mass. Now we assume that no other forces act or appreciably change these velocities as and when they collide. So linear momentum can only be changed due to collisions and the only expendable energy is the kindinetic energy with respect to center of mass.

[00:28:35]So for our case this energy of the colliding nuclei is essentially the kinetic energy with respect to center of mass. In reality at speeds close to speed of light conservation of linear momentum does not work. There lies a deeper concept which you can research about.

[00:29:02]Now energy is greater than the coolum barrier yes the fusion takes place with no hesitation but for energy less than coolum barrier in a classical sense you think yeah they'll stop and no fusion take place but when one studies the laws and the absurd absurdities of quantum mechanics you will understand that these things are possible even at energies less than the coolum barrier.

[00:29:36]George Gamour, a scientist who proved that particles can tunnel through a barrier even when possessing less than the barrier energy and through his gamma penetration probability. He proved the probability of a particle penetrating through the barrier and essentially reaching the other side.

[00:29:58]Now we'll see how did he do that.

[00:30:02]Suppose the out outside the target nucleus they consider a stationary nucleus one and a particle is approaching 2. Now the kum interaction potential is essentially V of R equals to Z1 into Z 2 E² by R.

[00:30:27]Now their relative kinetic energy is less than the coolum barrier energy.

[00:30:34]So it should stop. But gamma tunneling probability says no. It can go. It can not it must.

[00:30:44]In order to find out the probability.

[00:30:47]First we solve the radial shinger wave equation where s represents the particle wave function. the particle which is coming towards the target nucleus. Here mu is the reduced mass system. Essentially we are considering only two particle colliding. So we take the reduced mass of the system. We take the substitution s of r equals to radial distance into capital r of r.

[00:31:16]On rearranging we get the following equation.

[00:31:21]Upon some mathematics and few other stuff we get the solution s of r equals to exponentially growing term plus exponentially decreasing term. Now the exponentially growing term is discarded because the chance of the particle existing in the forbidden region itself is very low. It cannot always be positive like it cannot always be infinitely growing.

[00:31:54]So the exponentially growing solution is unphysical and discarded thereby giving us the only exponentially decreasing solution.

[00:32:05]Here kapa is defined as square root of 2 mu * v minus e by h cross².

[00:32:15]Now from your atomic structure those who has studied + 1 +2 might remember that probability density of a wave function is proportional to the modulus squared of wave function.

[00:32:31]So for our case it is exponentially decaying function. So let us consider it is equal to some e power minus g. Now upon squaring and writing it down we get e power minus 2g where g is equals to integral of capa dr r. dr is the small radial distance change. Basically we are calculating in a spherical system not in the xyz coordinate system.

[00:33:00]Now for the integration to take place we require the limits. The limits which we take are from the point at which the particle stops because it's kindinetic energy has fully converted into potential energy and it is not sufficient to cross the barrier. So it stops at a point greater than the barrier height and beyond the nuclear radius.

[00:33:29]Now that will be our outer limit. The inner limit would be the target nucleus radius which is essentially the nitrogen nucleus radius.

[00:33:42]So from the nuclear radius to the point where coolum potential equals to relative energy we integrate.

[00:33:52]We find this some sort of a weird integral which can be simplified using the substitution r = r2 sin square theta. Upon approximations we arrive at the final result g = to roo<unk> of 2 mu * e by h cross into pi r2x2.

[00:34:13]Upon further rearrangements and all we find the gamma tunneling probability is equals to P the probability equals to E power minus 2 pi EA where EA is equals to Z1 Z2 E² by H cross V.

[00:34:30]V is the relative velocity we talk because we consider one ed stationary with respect to one the other is moving towards it for the nuclear fusion to take place.

[00:34:46]Now we have learned the fusion can take place regardless of the energy. If the energy is greater than the coolum barrier, yeah it's it can happen with no hesitation. If it is less than coolum barrier, it might happen.

[00:35:03]We basically we got to know the probability of it being a successful reaction.

[00:35:11]Now using that probability we can find out how many reactions are taking place in a particular sample. Now how would we do that? Imagine you are traveling in a car. The chance of you crashing onto another vehicle is some P. Now the amount of times you crash into other vehicles depends on the density of the traffic. If it is a heavy traffic road, the chances of you colliding with into other vehicles is higher.

[00:35:44]The probability might remains the same, but the density has increased such that there's now an increased number of collisions that can be possible.

[00:35:56]If the traffic itself is very low, you probably won't hit many or none at all.

[00:36:04]Similarly, reaction rate is also calculated in a way the inter number of particles per second times the probability of a hit.

[00:36:16]Now, previously we came across this term cross-section area. Cross-section area in nuclear physics is an essential measure of reaction probability. The more the cross-section area, the higher probability that the reaction take place and its units are 1 bar= to 10^ - 24 cm²ared.

[00:36:40]For our case, the nitrogen nitrogen fusion reaction. There was a slight problem. There was no data for this collision reactions.

[00:36:49]Why? Because primarily the particle accelerators were not that capable enough to study these reactions.

[00:36:57]And secondary this was not the primary focus of the research. The focus was mostly on developing uraniumbased atom bombs, hydrogen fusion based fusion bombs and others. Nitrogen fusion was hardly of any concern.

[00:37:18]So since we don't have any kind of information on cross-section areas, the scientists began to assume as usual the best worst case scenarios.

[00:37:32]They took up took forth two types of models. The first being the constant cross-section or else every collision is effective. Basically every collision possible will result in a nuclear fusion reaction.

[00:37:48]But this is impossible. Why? It essentially states that you can ignore the barrier energy and it is it does not depend on the energy of the colliding nuclear itself.

[00:38:02]The assumed value of the cross-section is two bonds.

[00:38:08]But why did they do it? as usual the best worst case scenario.

[00:38:14]Now one point will strike to your mind or I'm striking it to you that why did they choose two bonds only if you calculate the radius of the nitrogen nuclei using the theoretical formulas you will find out it is approximately equals to 2.9 into 10^ -3 cm. Upon using this theoretical values and calculating the cross-section areas, we receive that it is 1.1 bars.

[00:38:49]So in order to overestimate the assumption and provide the best worst case scenario, they intentionally assumed it to be two bars rather than the 1.1 which was calculated from theory.

[00:39:03]This means it changes a lot the calculations and the severity.

[00:39:10]Now the sec second kind of model the scientist adopted is the not constant cross-section or not every collision is effective the reality if the energy is greater than or equals to the coolum barrier. Yes it definitely take part in the nuclear fusion reaction essentially uh contributing to the atmospheric ignition. But if the energy is less than the coolum barrier, we apply the gamma probability the tunneling probability where we multiply the constant cross-section which we assume the two bonds into the probability. Now you would ask the why is the cross-section itself changing? It doesn't make any sense. It will make sense. Now the cross-section is an indicator of reaction probability but the reaction probability itself is dependent on whether the energy is sufficient enough or not.

[00:40:15]So if the energy changes the cross-section should also change.

[00:40:20]Essentially that's the reason we adopt this form. when the energy is less than the coolum barrier.

[00:40:31]We have done with the with the collision cross-section assumption. Now we have to find out how much energy after collision it releases. We have seen the five types of reactions which are possible. But you would ask which like how many are taking first reaction, how many are paths are second and all. We don't know.

[00:40:57]At the time the scientists didn't know many different reactions are possible, different values. It's an absolute chaos. So they assumed only 17.7 mega electron volts.

[00:41:10]Why? because the ratios of different reactions the probability of ratio of ratios of different reactions it was not known at the time and the 17.7 is the maximum energy produced I asked you to remember the first reaction radiative capture which released even more greater than this alpha emission reaction you would ask why didn't they choose that because it releases gamma rays gamma rays take take almost the energy produced and they essentially escape the region of interest. They are essentially the atmosphere is transparent and cannot capture these gamma rays.

[00:41:52]So as always considering the best worst case scenario we assume the energy to be 17.7 mega electron volts per reaction.

[00:42:06]Now we are diving into more mathematical parts. So feel free to take screenshots or maybe ask charg after the session.

[00:42:19]Now the assumptions which we done the cross-section models the reduced mass systems relative velocities energy value assumption. Now we come to the number density assumption. Essentially the reaction depends on how many are present, how many nuclei are present.

[00:42:35]More the marrier because more nuclei will lead to more reactions and more catastrophic results.

[00:42:44]The scientists assume the number density to be 4 into 10 ^ 19 per cm.

[00:42:51]Now why that specific value?

[00:42:55]They give a time scale argument. The nuclear explosion generally take place with order times of order 10^ - 10 to -12 or 10^ -2 to - 10 while the expansions itself take around like 10^ - 8 seconds.

[00:43:15]Now these are very small when we try to observe these changes are very very tiny small but when you compare both the expansion of the gases itself takes more time.

[00:43:28]So by the time the expansion take place and the pressure changes and the volume distribution take place the explosion already would have happened that energy would have transferred already. So in such short span the energy is given to the nitrogen nuclei. So essentially the number density does not change appreciably in such time scale.

[00:43:53]Another thing which was you utilized in the calculations was the Maxwell Bsman distribution.

[00:44:01]You you would have studied this in your plus one plus2 syllabus about this and it's information about particles having more energy less energy and most probable energies.

[00:44:16]For our case we'll be using this equation because we are considering two particles and two particle system includes the reduced mass system.

[00:44:30]and the relative velocities.

[00:44:34]As a part of this session, I'm giving you a task to find out why do atoms possess velocities as per Maxwell Boltzman distribution at thermal equilibrium. Why not any other distribution? If you find out any other distribution, you could give it your own name.

[00:44:55]In case you forgot the reduced mass system is essentially m1 into ms2 by m1 + m_sub_2. We have completed the energy per reaction. Now energy rate energy rate is equals to reaction rate into energy per reaction. Like how many reactions are taking place and how much energy is being released. We have assumed the energy per reaction. Now we have to find out the reaction rate. Now how do you find out the reaction rate?

[00:45:25]Let us look at a small you would say a past look towards kinetic theory of gases.

[00:45:36]The kinetic theory of gases in assume two partic two species of particles with number densities n1 and n_sub_2. Now a particle with speed v is approaching the species 2. Now in some small time it will sweep a volume of v dt * sigma. Why sigma? Because it is the essential cross-section area. Basically around this wall around this area only the collision can take place.

[00:46:11]So about this volume is covered by just one particle.

[00:46:16]Now in order to find out the number of species 2 present in that volume, we'll multiply n_sub_2 with the volume sweep.

[00:46:28]Now this is dn = to n_sub_2 * sigma vdt.

[00:46:38]Now in order to find out how that one particle from species one meets the rate at which it meets the species two particles we'll divide dn by dt.

[00:46:51]Essentially we are treating the operator as fraction but at dt tending to zero we we'll get this n_sub_2 * sigma * v basically rate of reaction per particle. Now we would ask for whole particles simply multiply it with n_sub_1 because n_sub_1 is the density of particles present thereby we get the collision rate density but the above formula is not applicable why the cross-section changes the velocities are not same we assume that the whole sample follows the maxwell bsman distribution hence not every particle possesses the same velocity Therefore we do we do the average averaging thing where we average the product of cross-section into the velocity over the entire velocity distribution.

[00:47:54]Now this average is also known as thermal average.

[00:48:00]In order to find out this thermal average, we'll be using our Maxwell Boltsman distribution.

[00:48:08]The probability density function which is specified here.

[00:48:14]This FB into DV essentially represents the fraction of molecules having speeds between V and V plus DB where FB is the probability density function.

[00:48:31]We have to we have two models of cross-section. Hence we have to do two different calculations.

[00:48:38]The general formula for average is given here. Integral of zero integral from 0 to infinity sigma as a function of velocity time velocity time the probability density function for a small state interval.

[00:48:59]Now for the constant cross-section assumption, the constant cross-section assumption is essentially as the simplest one.

[00:49:11]So the sigma itself is constant. So we can take it out of the integration and integrate this v * fv dv from 0 to infinity.

[00:49:21]Upon doing the calculations, as you can see, the sigma is taken out of the integral and we find out that this whole integral or value will come out to be the average velocity of the sample.

[00:49:40]So the result corresponds to two two times the average velocity of the sample.

[00:49:50]Upon putting it in the energy rate formula we get that energy production rate is directly proportional to square root of t. So as temperature increases the energy production rate increases.

[00:50:08]Now for our variable cross-section assumption for energy greater than or equal to colum barrier we have done the previously the calculation which is essentially the constant cross-section assumption but for energy less than kum barrier we have to use the gamma tunneling probability essentially it's a variable cross-section we define sigma to be as 2 * e power minus b by v where B is defined as this 2 pi * Z 1 Z 2 E² by H cross.

[00:50:45]Upon pulling out all the constants and keeping the variables inside the integral we get this V cub * E minus B by V and E power minus mu V ² by 2T * DV.

[00:50:59]Now this is a very complicated integral.

[00:51:02]If you want you can try. Those were math nerds but And no elementary anti-airway to exist for this above integral.

[00:51:12]So they have to approach through different methods of integration.

[00:51:17]The numerical methods the exponential consists of two terms.

[00:51:23]One due to Maxwell Boltzman distribution one due to gamma probability.

[00:51:29]Now if the velocities are very very low this exponential term since V is in the denominator as V is decreasing the whole part increases and basically the low energy particles cannot cross the energy barrier whereby the only possible reaction taking place is through tunneling.

[00:52:08]So the gamma factor dominates the than the Maxwell factor the exponential.

[00:52:18]Now keeping this in mind at low temperatures a sub case in the variable cross-section at low temperatures the reaction is possible only through tunneling.

[00:52:30]So we take this substitution x= to b by b and calculate its derivatives as well and plug them into the integral. we get this sort of integral x^ - 5 * e^ - x e^ - mu b² by 2t x² dx. Now e power x falls off very rapidly compared to that other exponential.

[00:53:02]So the other exponential essentially almost remains the same. So we take an approximation of the other exponential and using the tailaylor expansion which you have stud which you have studied in 11th class limits e power minus y is equ= to 1 - y + y² by 2 factorial and so on where y is this value we approximated to only second term because beyond which the contributions are neglected.

[00:53:36]Now on these two terms are plugged in and the calculation is conducted we get that now this calculation is done using the gamma function which is defined as gamma of n is equals to integral of 0 to infinity x^ n -1 * e^ - x dx.

[00:54:01]Now using this gamma function we calculate and the thermal average comes out to be t power proportional to t power minus 3x2 * exponential exponential time exponential of minus constant by t.

[00:54:21]As temperature increases, this denominator becomes a huge value.

[00:54:26]Thereby this whole fraction becomes a smaller quantity.

[00:54:32]So this exponential increases.

[00:54:36]So as temperature increases there's a much higher probability of the particles fusing and conducting in taking part in the fusion reactions.

[00:54:48]These calculations are conducted for less than 0.2 mega electron volts temperatures.

[00:54:56]Why? Because the data shows that at such low temperatures the energy possessed by them is so very low that only for means there is that is through the tunneling probability.

[00:55:13]Now at the intermediary stages where beyond the 2 mega electron volt temperature now the exponential terms which we which consists of these two let us consider them as a function or the whole function. So we'll get this.

[00:55:31]In order to calculate here we use method of steepest descent. Instead of integrating the whole thing we'll find out where the maximum contribution comes.

[00:55:43]Why we do this? Because the area under the Maxwell distribution and the area under the gamma probability distribution overlap at these points and give us a certain peak where the maximum contribution take place.

[00:56:05]So we'll try to find out that peak itself the majority contribution.

[00:56:10]We differentiate this function which we have defined and set it to zero to find its minima.

[00:56:17]This velocity is known as gamma peak velocity.

[00:56:20]And using tailor expansion series around the v kn of v as in terms of the tailor expansion series. Notice that the first derivative is missing because at first derivative V not it's at first derivative we at V not it becomes zero.

[00:56:42]Now using this scalar expansion we'll plug this in into this exponential essentially approximating the integral and we take out the constants out and we get this kind of integral which is a gshian integral.

[00:57:00]This gian integral has a result which is from minus infinity to infinity e^ minus ax² dx gives result by 8. Upon using the result and plugging in the values of 5 n ddash we get that the thermal average is proportional to a * t power - 2x 3 exponential of - 3 * B ² mu by 4^ 1x 3.

[00:57:33]Now these a and b are numerical constants which are to be determined by the the constants pi values the cross-section values other now but is is the nitrogen only the suitable candidate? The scientists asked could there be any other possible candidate that they they would have missed.

[00:58:05]They began searching and they got an idea. What if the bomb is detonated near a sea?

[00:58:12]It would release tremendous amounts of water vapor thereby raising the probability of hydrogen nuclear protons.

[00:58:20]Now they conducted analysis on this protons and whether they could help in the reactions.

[00:58:28]And what they found out is hydrogen nuclei are definitely not sufficient to take part in the nuclear reaction. Why?

[00:58:38]First of all, the reaction cross-sections itself were very small compared to compared to that of nitrogen nitrogen nucleus.

[00:58:46]And secondly, the reaction energies which were produced were very less compared to that of nitrogen nitrogen.

[00:58:54]So essentially they do not contribute much and the abundance is also a question that even in detonating near the sea in comparison to the 78% of nitrogen it is always less the hydrogen nuclei are always less than the nitrogen nuclei thereby the possibility of other candidates were ruled out after all this calcul ations they plotted a graph.

[00:59:28]This graph is the is ploted due to the constant cross-section assumption and the variable cross-section assumption. The blue line indicates the constant cross-section assumption line which essentially is directly proportional to square root of t. Now the variable cross-section notice it starts at some temperature and it it will go on and at very high temperatures they tend to meet.

[01:00:01]So these two model and produce the same results almost at highest temperatures.

[01:00:11]So essentially we learn how the energy is produced in a case of atmospheric ignition.

[01:00:18]But the production of energy is not sufficient enough. You need to understand how the energy is lost also.

[01:00:27]Now Olive is going to continue with the energy loss part of the session.

[01:00:41]Now that we have seen in detail how the fusion reaction occurs and energy is produced, let us go into energy losses.

[01:00:49]Before that, I would like to define something called as the safety factor.

[01:00:55]The main goal of this particular section is to see that if the rate of energy loss exceeds the rate of energy production and if it does we are safe.

[01:01:06]So we define s the safety factor to be the ratio of the rate of energy loss to the rate of energy production and we will check whether this s is greater than one.

[01:01:21]Now before we actually move into energy losses, one thing to note is that the explosion of the fision bomb produces high amounts of energy. As already mentioned, at these high temperatures, the extreme kinetic energy tells all orbiting electrons away from the nuclei.

[01:01:45]So instead of the atmosphere being filled with nitrogen atoms, oxygen atoms and so on, we don't have full atoms but we have uh what is called a plasma which consists of free electrons and nuclei.

[01:02:04]Yeah. And if you've not heard of the term plasma before, it is the fourth state of matter that exists when there is very high temperature. Some examples of plasma include our sun, stars and even lightning.

[01:02:23]And uh one thing that I would like to add over here is that because the temperatures are very very high it becomes difficult to keep using Kelvin. So what I will you do is multiply the temperature by the boltsman constant which is given on the screen to convert the temperature into uh me units where 1 me is um corresponds to 1.16 into 10 ^ 10 kel.

[01:03:00]All right.

[01:03:04]Now that um we have established that our atmosphere is in the state of plasma.

[01:03:12]We go back to the equation here. The the nitrogen fusion producing magnesium and alpha particle giving 17.7 mega electron volts of energy.

[01:03:28]This energy is in the form of kinetic energy that the product particles have.

[01:03:35]Now how this energy is transferred is uh mainly by collisions. So these uh particles are not static in the atmosphere. They keep moving about randomly and they collide with each other and that is how energy is transferred from the product particles to the surrounding particles.

[01:03:55]Now since the mass of an electron is much much smaller than the mass of the nuclei energy is much more rap uh rapid between particles of comparable mass.

[01:04:09]So the energy from the product particles is first shared to all the nuclei establishing a nuclear temperature and electrons because of their very low mass receive this energy after many more collisions only and therefore uh the electron temperature is lower than nuclear temperature.

[01:04:40]Now the rate at which the nitrogen nuclei transfer energy to electrons.

[01:04:48]The atmosphere has not only nitrogen nuclei but also oxygen and other nuclei.

[01:04:53]But I will be considering only nitrogen as it is the most abundant nucleus.

[01:04:59]And um we have this very uh fancy looking formula that is calculated assuming only the collisions of the nitrogen nuclei with the electrons. So the rate of this energy transfer is given as 4 pi * n. n is the number density of nitrogen in the atmosphere.

[01:05:21]Zed in in this case would be the atomic number of nitrogen. M is the mass of nitrogen nuclei. E is the charge of an electron and we have a TE which is the electron temperature and TN which is our nuclear temperature.

[01:05:38]Now as you can see here this uh rate of energy exchange is proportional to the average of uh reciprocal of velocity. The bar above the one by V indicates average.

[01:05:54]As uh Satik already mentioned uh to to see how the velocity is distributed at a particular temperature, we generally use what is known as the Boltzman distribution. And in the Boltzman distribution, our velocity a uh reciprocal velocity average would be root of 2m by pi * t electron.

[01:06:19]But our atmosphere uh this region is in a state of plasma high with high amounts of energy high temperature right and so uh also uh one thing to note is the velocity here is the velocity of the nitrogen nuclei because they collide with the electrons and so these nitrogen nuclei they travel at huge amounts of speed due to the high temperature almost close to the speed of light. What happens when you start moving near the speed of light?

[01:06:55]Relativistic effects start uh popping up. And to account for that, instead of the Maxwell Boltzman distribution, we use something called as the Maxwell Utner distribution. The name Utner is German, hence the pronunciation.

[01:07:13]Um so according to the Maxwell utner distribution we uh have an equation that gives us the average to be 1x c. c is the speed of light* 2x pi theta. We define theta to be uh electron temperature divided by mc² and um as mentioned already when I whenever I say temperature I am using temperature in units of energy. So theta over here is a dimensionless constant and you have uh the numerator depends only on theta while in the denominator you see something called an h.

[01:07:59]This h denotes hankl functions. This is beyond the scope of this session. I have only mentioned it here because uh it will come up later. But you will see that these hankle functions cancel out later on. If you're interested, you can always u I would suggest that you go and look up max distributions.

[01:08:24]All right.

[01:08:27]Now that we have established that the atmosphere is in a plasma state and you have free electrons and nuclei, what do you think happens?

[01:08:39]Electrons have negative charge and the nuclei are basically made up of protons and neutrons. Neutrons of course have no charge whereas protons have positive charge and uh if you know already opposite charges attract. So these electrons would be attracted towards the positive charge proton nuclei.

[01:09:08]Okay.

[01:09:10]So our high energy electrons move towards the nuclei and as they move towards uh the nuclei distance decreases which makes the force increase and the electron starts to accelerate causing it to bend around the nucleus.

[01:09:31]Now it is important to note that the electron is not directed straight towards the center of the nuclei but at a perpendicular distance and I will call this perpendicular distance impact parameter denoted by B and I will come back to that in just a minute. Right? So uh so our electron is attracted to the nuclei and as it moves closer because it's not directed uh to the center of the nuclei it bends around the nuclei so that it can due to the force kimic force right now according to electronamics whenever a charged particle accelerates it must radiate energy and because our electron is changing direction it is essentially accelerating and this uh radiation the power of this radiation is given by what is called as the law formula where P denotes the power of radiation which is equal to Q ² * A² by 6 pi epsilon * CQ is the charge of the particle that is accelerating a is its acceleration and C is the speed of light uh And this radiation is given off in terms of photons. So this uh wavy line coming out of the nuclei is essentially the photon that is transmitted out. And obviously this radiation has some energy. But then this energy cannot be created out of nowhere. Right? And because of this radiation our electron uh electron's kinetic energy reduces that is this electron slows down.

[01:11:21]Hence the and this phenomenon of an electron slowing down is called as bremstral radiation. Bremstung is u means breaking. The name came because electron basically slows down or like hits the brakes right. So it is known as breaking radiation.

[01:11:46]And um remember I mentioned I will circle back to impact parameter and I said that the electron is not uh aimed directly at the center of the nucleus.

[01:11:59]But if it was aimed at the center of the nucleus, it would be a problem and it it may not even be aimed exactly the center of the nucleus. If the impact parameter was small enough or as it tends to zero the radiation would tend to infinity and obviously we cannot have infinite radiation. It becomes a problem.

[01:12:22]This is where quantum mechanics helps us.

[01:12:28]Electron over here is not just a particle but also a wave. Right?

[01:12:36]So uh and we also use Heisenberg's uncertainty principle that tells us that um you cannot know both the position and the momentum of the electron at the same time using that and uh using that we say that the minimum value of this impact parameter has to be the D broadly wavelength lambda here is the D broadly wavelength which is calculated as h bar by mu. H bar is a reduced planks constant and mv is the momentum and uh using all of these formulas lamos formula kombic interaction formula and this you calculate this uh formula for the rate of energy loss due to brastston.

[01:13:31]Now one thing a side note sort of that I would like to mention is that BMT radiation is a continuous spectrum as uh mentioned already because of the existence of minimum value of the impact parameter there is a maximum value that the radiation via bremstronum can take take right but is there uh say like only so this radiation can only take certain values No, >> because our electrons are free. They're not orbiting the nuclei. They they're free. They can go anywhere they want.

[01:14:10]These electrons uh can take any values.

[01:14:15]The velocity of these electrons can take many values. And therefore the energy difference is a continuous uh spectrum that goes from zero to this maximum caused by impact parameter.

[01:14:31]And u drilling radiation is a continuous spectrum.

[01:14:36]Now coming back to this formula uh so we have said that the rate of energy loss via bstling radiation is equal to 16x3 * n z where n is the uh number density zed is atomic number multiplied by z square e again is charge m is mass of electron cq and h bar and you have uh the average of B * 1 + E by MC².

[01:15:10]And you might wonder what is this 1 + E by MC²? Where did that come from? Well, as mentioned already, this uh whole event occurs in a plasma region. There is a huge temperature and we have the electrons moving really really fast. Right?

[01:15:33]So what we have to do uh is use something called a lorren factor.

[01:15:40]this and 1 this 1 + e by mc² is essentially the lurens factor and we multiply that with the velocity and then we calculate the average velocity using the maxwell distribution again and we get this fancy equation where again theta is t by mc² and you have the this hankle uh function in the denominator.

[01:16:09]Now if you noticed the Hankle function in the denominator of the uh nitrogen nuclei to electron energy transfer and the Hankle function here is the same right that will help us in a little bit of time and as Satik had already mentioned the average velocity by Maxwell Boltzman distribution would be roo<unk> of 8 T by PM And if we consider the uh temperature to be very low and u remove all relativistic effects then uh this 1.7 times root of electron temperature would be the rate of energy loss.

[01:17:02]So now our main goal of this session was to uh calculate the safety factor right but if you had u seen previously in the slide that Zik had shown he used the nuclear temperature to calculate the uh rate of energy production whereas almost everywhere here to calculate the rate of Energy loss via bremstral we use the electron temperature. Okay, this velocity depends on the electron temperature. So we try to uh see if we can find a relation between the nuclear temperature and the electron temperature. And to do that what do we do?

[01:17:46]We assume that the rate at which the electron loses energy to brimstone radiation is the same rate at which the electron gains energy from the nitrogen nucle. And when we equate that we get this really nice relation between uh nuclear temperature and electron temperature which is only dependent uh on electron temperature and MC².

[01:18:15]And you have this graph that shows us electron temperature with respect to nuclear temperature.

[01:18:29]And then we have the graph on the left where the yaxis represents rate of energy change and the xaxis gives us temperature. And as you can see as the temperature increases the rate of u energy production and the rate of energy loss by bremstrom they're almost the same towards like the very right of the x-axis right and from that when we calculate the safety factor we see that the safety factor reduces with nuclear temperature and we reach an empirical minimum of 1.6 6 at a nuclear temperature of 100 mega electron volts.

[01:19:13]Now obviously 1.6 is not ideal. Yes, it is greater than 1, but it's not as good as we would like.

[01:19:24]But then if you look at 100 mega electron volts of temperature, that is about 100,000 times the temperature of our sun's core, which would make you think, oh wow, that is good. But um when it comes to the end of the world, we have to be very pessimistic. So we will look at other energy losses too.

[01:19:49]And as mentioned earlier, we have a plasma region, right? And we've got these u high energy electrons in this plasma region and also photons that are produced by our branching radiation. Now under normal circumstances these photons would just get radiated away. But if you consider uh the volume of our plasma region to be big enough then instead of just getting radiated away these photons uh start colliding with these high energy electrons and when they do that u the high energy electrons uh lose energy and photons end up gaining energy. This effect is known as the inverse Compton effect. Why inverse Compton effect? Because there also exists a Compton effect where u high energy photons are scattered by low energy electrons.

[01:20:52]Now because these photons are produced due to bstling radiation there exists a relation between this loss due to inverse compent effect and loss due to bstling uh radiation which is given by 10x3 * electron temperature divided by mc² into r * lambda. Now what is lambda?

[01:21:14]lambda over here is the component mean free path. And if you don't know what a mean free path is, it is basically the average distance that a moving particle travels between successful collisions.

[01:21:27]In this case, it would be the distance that a high energy electron would travel before it uh goes and hits a photon. Right? And this uh content mean free path is known to be 42 m.

[01:21:45]And because um this inverse Compton effect is directly proportional to the radius of our um plasma region.

[01:21:56]We will um now see if um in what is the optimal value of this plasma region should be. Now you might wonder why do I say radius? Why couldn't the plasma region be say a cuboid or any other shape? Well, uh we assume that this region is a sphere because of symmetry and to make the lives of physicists easier.

[01:22:25]Now, uh the next energy loss would be uh we've we've said that there is a plasma region, right? But then the air surrounding this plasma region would be at normal temperature and then the energy would uh be transferred to the uh volume outside due to collisions.

[01:22:53]So when particles inside this plasma undergo like random motion and they collide with each other and when they leave this region and collide with a particle outside they transfer energy to the surrounding air.

[01:23:11]And so for our nitrogen fusion reaction to be self sustaining a new fusion reaction must occur before the energy can be diffused away. Right? And that is what we will consider in this section.

[01:23:26]Now what we will do is uh even though this uh plasma region has photons, electrons and nuclei, we will consider photons to uh diffuse the energy outside because most of this diffusion occurs by photons and u the photons just don't just move from inside this plasma region sphere outside in a straight line. they collide within the sphere itself and after multiple collisions only they move out.

[01:24:00]And when you take the r to be the distance that the photon actually travels before it moves out. Then you can calculate d escape which is the actual uh distance that the photon would travel after multiple collisions.

[01:24:21]And you have this formula that gives us dsk equals to r² by lambda where lambda again is the mean free path between uh collisions that is the average distance between two consecutive scattering collisions that happen within the plasma region itself. And we calculate d fusion which is the fusion mean free path which is calculated as 1 by n * sigma where n again is the number density and sigma is the cross-section.

[01:24:57]Uh so essentially we uh we equate this d escape to the d fusion to calculate what the minimum value of radius should be. And when we do that for the nitrogen fusion reaction giving magnesium and alpha particle we find out that the radius is 57 m.

[01:25:23]Now to produce u a plasma region with say uh with say a nuclear temperature of 10 mega electron volts. I'm using 10 mega electron volts instead of 100 mega electron volts because if you had noticed in the graph earlier the safety factor at 10 itself was uh two right and 100 is way too high for us. So we assume 10 mega electron volts and when we do that we find out that we will need 1.5 * to 10 ^ 6 kilogram worth of material to create such a plasma region.

[01:26:08]This obviously is too high. So then we try to see if we can somehow reduce this radius.

[01:26:18]Now uh the lambda used in ds escape is obviously a constant and number density is also a constant. So the only way that we could reduce radius is by reducing this uh sigma.

[01:26:33]So in the second equation that I have given here what happens is nitrogen fuses to form oxygen and carbon and in this u reaction the nitrogen nuclei do not actually need to come into contact. One nitrogen nucleus splits off uh dutium nucleus that then attaches to the other nitrogen nucleus forming oxygen and carbon nucleus. Right?

[01:27:02]and we see that the radius from such a reaction would be 7 m.

[01:27:09]Is this a problem? Well, no. Because the amount of energy that the second reaction gives is only 10.6 me. So, our safety factor when multiplied is uh increased by a factor of 1.67.

[01:27:28]And with that we come to the end of the energy losses.

[01:27:36]But as I've mentioned here, is it possible to create a bomb that has this huge radius 57 m and you know that could actually ignite the atmosphere. Sankit will take over from here.

[01:27:58]Yeah. So till now Olive and Satik has been speaking about the energy production as well as the energy losses that are involved in the nuclear explosion. So the next part of the discussion is going to be what if we could at least theoretically make a bomb that can set the air on fire.

[01:28:19]So before we go into that, let us go back to some thermodynamics which you would have done in your high school.

[01:28:28]Suppose that you have a fixed amount of energy Q. Why a fixed amount of energy?

[01:28:33]Well, even if a nuclear bomb releases a very huge amount, it is still limited because it depends on the chemistry of the nuclear material that is being used.

[01:28:42]So your energy even in this case is still finite. The amount the temperature change that can be produced with a fixed amount of energy is inversely proportional to the quantity of the matter that is being heated is inversely proportional to the quantity of matter that is being heated which is quite clear from this relation given here. So the temperature change is inversely proportional to the volume being heated. Now you may think that if we minimize the volume well the temperature change that can be brought about by a fixed amount of energy it can be increased right as a denominator goes down delta t here goes up. Yes that may seem obvious but actually that is quite a n conclusion to arrive at. That is because the propagation of the reaction demands that the energy production in each newly entered region exceeds the losses from that region.

[01:29:43]What does this mean? Well, let me break this down for you. See, when a nuclear reaction happen, energy is produced as the nitrogen nucleic collide and fuse with each other. This energy, well, it is lost primarily through bremstral as well as other modes such as inverse compent effect. The energy that is produced it is actually carried away by the product particles primarily the alpha particles who have an initial kinetic energy of appro 17 mega electron volts. These alpha particles they travel outward from the reaction site and gradually deposit their energy into the surrounding medium. What do you mean by this deposition of energy? Suppose a fast charged particle uh in our context an alpha particle with about 17 meal electron volts of kinetic energy. It is fired into air. Now it does not travel through the air like a bullet fired into empty space. Rather it interacts with the other particles around it.

[01:30:45]Now it constantly feels the electromagnetic forces that are acted upon it by the other particles and these interactions slow it slows it down.

[01:30:57]However, is this loss uniform or does it follow some pattern? This actually determines our argument of whether the temperature change can be attained by heating a very small volume.

[01:31:13]See initially when the alpha right after the reaction when the alpha particles are produced they have a very high kinetic energy but when the alpha particle is moving very fast it moves past each electron in a very small fraction of a second the electron barely has any time to interact with these alpha particles and respond. Hence a fast particle loses energy slowly per unit distance. Now as the alpha particle moves it slows down because its interaction with the surrounding particle increases and it loses energy more rapidly per unit distance.

[01:31:51]If you look at this graph you can see that initi this graph here represents the rate of loss of energy rate of loss of energy of the particle as it moves through a medium.

[01:32:05]This is called the Brack curve named after the British physicist William Brack who did the calculations and arrived at a result in the early 1900s.

[01:32:14]On the y-axis we have here the stopping power which is nothing but the energy released per unit distance. Well, just after the reaction, the alpha particle has a very high kinetic energy. However, the energy that it releases per unit distance, it's low and roughly constant.

[01:32:32]As it proceeds this increases and reaches a sharp at a particular distance and then it comes to rest as it turns into a neutral helium atom. This is because at different intervals the time with which it interacts with other particles is different. This introduces us the concept of interaction time.

[01:32:55]The interaction time is basically the time that a moving particle spends near another particle such as an electron or a nucleus. As far as we are concerned, the physical quantity that matters to us is the path length. That is the distance that the alpha particles move before they come to rest.

[01:33:24]So this distance is actually called the mean free path or the range of the alpha particle.

[01:33:31]Now this so the conclusion that we have arrived at is that the alpha particle does not actually deposit all of its energy right at the reaction site rather it moves a particular particular distance and gradually deposits all of its energy deposits its energy and most of it is deposited at a particular distance away from the reaction site. So our initial argument was that if the volume is brought down extremely small the temperature change can be huge.

[01:34:04]However the ar however that argument is not valid because the energy is not actually deposited only at the reaction site. So our question changes to how much energy does it take to heat a sufficiently large volume to the required temperature. In order to know that we need to understand how big that required volume is. For that we need to find out the value of the range that is lambda here which is equal to 1 upon n * sigma total. Here capital n denotes the density of particles in the air and sigma total is the cross-section area that the particles present for the interaction.

[01:34:48]Now to calculate that we need to find out some relation that would give us the total area of cross-section. So when the alpha particle passes an electron uh the energy of the energy transferred to the electron in a single collision it depends on the impact parameter B here and also the another fundamental result in the charged particle scattering in the RA is the RDA for cross-section that is when a particle of charge Z E passes near an electron of charge E the differential cross-section for scattering ing into solved angle that is d sigma by d omega is given by this equation and also the energy loss per unit distance. It can be found out from these results which comes from uh plasma physics actually and underlying beneath all this is another uh concept called the fer plank equation and der and by using some other relations from thermodynamics we finally arrive at a particular equation for finding out the area of cross-section. Now I would like to motivate you guys to read more about plasma physics and the equations that are actually involved in this derivation for the area of cross-section.

[01:36:19]So finally we arrive at this equation where sigma E that is the total area of cross-section for colombic interaction between the alpha particle and the electron turns out to be this particular equation. Now as Satri mentioned earlier what we are considering is the best worst case scenario that is we are assuming that the energy takes that the explosion takes place in the worst possible manner and the maximum energy is released for that the area of cross-section also needs to be maximized. So we differentiate this equation and set it to zero. From that we get that the relation is maximum when E is equal to 9x2 * TE. Here E is the energy of the alpha particle and TE is the temperature of the electron in the air. So the maximum value of the colombic interaction between the alpha particles and the electron is given by this relation. Now apart from this the alpha particle is also undergoing nuclear scattering because of the nitrogen nuclei that is present in air.

[01:37:21]This also provides a area of cross-section which is approximated to about 1.2 bars. So the total area of cross-section that is in front of the alpha particle is about 3.3 bars. Now as mentioned earlier this is a bit exaggerated value but we are considering the best worst case scenario. So now we have the value of sigma total required to calculate the range. If we plug in this value of capital n and sigma total we will get that the range of the particle that the alpha particle is about 57 m. What this means is that the alpha particle travels an average distance of 57 m before it actually deposits most of its energy.

[01:38:15]So its energy is not actually deposited at the reaction site itself but rather at a distance of 57 m away from there.

[01:38:24]What does it imply? Well, it implies that you need to heat up a volume or sphere of radius 57 m such that the reaction is sustained and it propagates in order for the atmosphere to get ignited.

[01:38:41]However, for such a huge volume of air to be ignited, you would be requiring an enormous amount of energy.

[01:38:49]That volume of air would approximately have about 4 into 10 ^ 31 nuclei, 3 into 10 ^ 32 electrons. And in order to heat up that volume of air, you would be requiring energy of the order of 10 ^ 32 mega electron volts. But a fusion bomb produces a maximum of 5 into 10 ^ 26 mega electron volts of energy and that too only if the reaction is 100% efficient. Typically nuclear reactions have an efficiency of around 1%.

[01:39:21]And even then you would be requiring a mass of 1.5 into 10 ^ 6 nuclear material for all of this to happen. Now all of this is actually far from achievable. Hence the physicists arrived at the conclusion that it is quite impossible to build a bomb that could set the air on fire.

[01:39:46]But is this the only threat? Well, the answer is no.

[01:39:52]If you have watched the movie Oenheimer, you would remember this scene where physicist Stella proposes an idea of building a fusion bomb. That is a bomb made of heavy hydrogen dutarium.

[01:40:08]See a bomb powered by fusion on paper it releases more energy per unit mass than fusion. So why not proceed with that one? Well, the thing is that in a fish bomb, it is very efficient at depositing energy back to the surrounding such that the material keeps the reaction self- sustaining. But dutaterium nuclei they are lightweight low charged particles.

[01:40:34]They don't radiate efficiently when they collide or interact. So the energy from fusion reactions doesn't like it does not help in reheating the way which is done in fusion. Now another thing is that for an amount of 6.4 megalon volts of energy to be released per dutarium all of these secondary reactions need to happen. The probability that these reaction happen is actually quite low.

[01:41:03]Especially the last part where a return combines with the helium 3.

[01:41:08]And even if all of this were to happen, in order to heat up a required volume of air, you would be requiring a sphere of radius 3 m of liquid dutarium, which was not feasible. Hence, we closed that pathway as well.

[01:41:27]Now another possibility earlier in the session Satik had discussed about all the posit all the possible reactions that could happen. Out of this the oxygen carbon channel actually seemed quite a bit dangerous at first glance.

[01:41:42]Why is it so? Well, it was mentioned by Olive earlier that in this case their range is approximately 7 m. See I mentioned that the range for alpha particle was about 57 m. it drops down to about 7 m in the case of oxygen and carbon channel. What does this mean?

[01:42:00]This means that the volume that you need to heat up that also decreases by a factor of RQ that is 57 upon 7 the whole cube. So on first glance this may look catastrophic but it did not cause much of a worry. The reason for this are mentioned here. The first one being the colum barrier.

[01:42:23]For this reaction to proceed through this channel, it had to overcome a colum barrier of approximately 14 mega electron volt. Now to put this into perspective, the coolum barrier of the alpha particle path was about 7 to 8 mega electron volts. Now you can argue that well the coolum barrier can actually be overcome by quantum tunneling which is a possibility that satic discussed. But the thing is that the quantum tunneling probability it is proportional to e ^ roo<unk>^ b by t.

[01:42:54]Here the capital b is the term that is representing the colombic interaction.

[01:43:00]Oxygen and carbon have comparatively higher charges and hence the colombic interaction is more. Hence the quantum tunneling probability is reduced.

[01:43:11]The second reason for this path for not being such a dangerous one is that the energy released per reaction is smaller.

[01:43:19]The energy released per reaction is about 10.6 mega electron volt which is less than the 17.7 mega electron volt of the alpha emission pathway. This increases the safety factor by a factor of about6 making it to 2.67.

[01:43:36]The third one is the inverse component effect. It was discussed a bit in detail about detail by Olive and further clarity will be given by Satik.

[01:44:00]Yeah. So the component scattering we have seen the energy production rates the energy loss rates how the energies produced due to the fusion reactions how it is lost due to burning radiative losses there's another thing the Compton effect the Compton effect is through which the electrons high speeded moving electrons transfer their energy to the photon tons which in turn escape.

[01:44:33]Basically they lose out the energy.

[01:44:36]The inverse compound effect is essentially the inverse of it where photons transfer the energy to the electrons.

[01:44:43]Now this inverse component effect is very critical because inside the fusion region where the electrons are carrying the energy they can escape out the region and essentially transmit away the energy.

[01:45:01]Now the photon carrying the lost energy if it comes back and due to inverse component effect it transfers this energy back into this region it stays for quite long time then the losses can be minimized and the scientists worried about this possibility whether this inverse component effect could significantly reduce bsting radiations but when they found out and did the calculations The calculations resulted that the photons event when the high-speed photons gain very much high energies they essentially escape out of the region instantly. So the chances of them transmitting back into the region are very less. Thereby the inverse component effect though existing does not significantly reduce the bremstral radiation.

[01:45:59]Hence the first graph you see the green color line which represents the brimstone loss rate is always greater than these two production rates.

[01:46:10]This ensures that the reaction region does not reach to a point of ignition temperature or the nuclear temperature where it cascades into a chain self-sustaining chain reaction.

[01:46:26]Upon all these evidences, calculations, and assumptions, in conclusion, the LA602 report established that the Earth's atmosphere cannot be ignited by fish or a thermonuclear bomb, essentially making it a dream.

[01:46:45]Even after conducting such in-depth research, the scientists were still terrified of the probability of a near zero probability of an atmospheric ignition.

[01:46:56]The first atomic test, the Trinity test on July 16, 1945.

[01:47:04]Many scientists were still pondering over the question whether the atmosphere would ignite and the world would get destroyed.

[01:47:24]As you can see the world did not get destroyed and the first nuclear test was conducted very successfully.

[01:47:34]Hence by through the conclusion of the report we can sufficiently sleep at our homes that the atmosphere could not be ignited due to a mere atomic explosion.

[01:47:52]This report is also a kind of an overview of how scientists meticulously create the worst possible outcomes for disasters and try to overestimate their chances so that they can find out what in what ways they can reduce the the damage caused.

[01:48:18]Thank you for attending the session. We hope you learned something today and feel free to drop any questions in the WhatsApp group.

[01:48:30]You can scan the QR for the attendance.

[01:49:14]Once again, thank you very much for attending the session. We will be ending the live stream in 2 minutes.