The Manhattan Project

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by Bruce Cameron Reed


  As to the role of the bombings in the Japanese decision to surrender, the Strategic Bombing Survey report came to mixed conclusions:

  It cannot be said, however, that the atomic bomb convinced the leaders who effected the peace of the necessity of surrender. The decision to surrender, influenced in part by knowledge of the low state of popular morale, had been taken at least as early as 26 June at a meeting of the Supreme War Guidance Council in the presence of the Emperor … The atomic bombings considerably speeded up these political maneuverings … The bombs did not convince the military that defense of the home islands was impossible, if their behavior in government councils is adequate testimony. It did permit the government to say, however, that no army without the weapon could possibly resist an enemy who had it, thus saving ‘face’ for the Army leaders … in the atomic bomb the Japanese found the opportunity which they had been seeking, to break the existing deadlock within the government over acceptance of the Potsdam terms.

  The Manhattan Project had been staked on the idea of turning a recently-discovered, incompletely-understood physical phenomenon into a weapon that could destroy an entire city. Three years after it had formally been wagered, that gamble paid off in the skies over Hiroshima and Nagasaki.

  References

  [1] Source: http://commons.wikimedia.org/wiki/File:Atombombe_Little_Boy_2.jpg

  [2] Reed B C 2014 The Manhattan Project Physica Scripta 89 108003 figure 24

  [3] http://commons.wikimedia.org/wiki/File:Fat_Man_on_Trailer.jpg

  [4] http://www.lahdra.org/pubs/reports/In%20Pieces/Chapter%2010%20Trinity%20Test.pdf

  [5] http://commons.wikimedia.org/wiki/File:Trinity_Jumbo.jpg

  [6] http://commons.wikimedia.org/wiki/File:Trinity_Test_Fireball_25ms.jpg

  [7] http://commons.wikimedia.org/wiki/File:Trinity_shot_color.jpg

  [8] http://commons.wikimedia.org/wiki/File:Trinity_crater_(annotated)_2.jpg

  [9] From Reed B C 2014 The History and Science of the Manhattan Project (Berlin: Springer) (http://commons.wikimedia.org/wiki/File:050607-F-1234P-090.jpg, http://commons.wikimedia.org/wiki/File:Atomic_cloud_over_Hiroshima.jpg, http://commons.wikimedia.org/wiki/File:Atomic_bomb_1945_mission_map.svg, http://commons.wikimedia.org/wiki/File:AtomicEffects-p7a.jpg, http://commons.wikimedia.org/wiki/File:AtomicEffects-Hiroshima.jpg)

  IOP Concise Physics

  The Manhattan Project

  A very brief introduction to the physics of nuclear weapons

  B Cameron Reed

  * * *

  Chapter 6

  Effects of nuclear weapons

  Nuclear explosions can injure and kill human beings, start fires, and destroy even heavily reinforced structures, all at great distances from their point of detonation. For humans unlucky enough to find themselves in the vicinity of a nuclear explosion, the most obvious feature will be an intensely bright fireball and searing heat. When a nuclear weapon is detonated, some 70–80% of the energy is released as thermal energy in the visible, infrared, and x-ray ranges of the electromagnetic spectrum. X-rays are absorbed by the air within a few feet of the explosion; this heats the air and creates a brilliant fireball which can cause blindness and serious if not fatal burns. Depending on an observer’s distance from the explosion, the next noticeable effect is likely to be a shock wave caused by a front of compressed air which propagates outward from the explosion, caused by the sudden release of a large amount of energy into a small volume of space. Even if an observer is shielded from flash, heat, and shock, several other effects, invisible but potentially no less harmful, can come into play. Fissioning nuclei emit highly-penetrating gamma rays, which are even more energetic than x-rays and can penetrate through air to great distances. Neutrons are estimated to carry off only about 1% of the energy liberated in a nuclear explosion, but are also energetic enough to travel through the air for hundreds of meters and cause cellular damage to persons exposed to them. Over longer terms, the neutron-rich nuclei which are the products of fissions decay to stable isotopes, primarily by beta-decay, which leaves the atmosphere and ground in the area of the explosion radioactive long after any immediate effects have been felt. All of these phenomena manifested themselves at Hiroshima and Nagasaki. This chapter examines these effects in the order listed here.

  Readers interested in analyses of the effects of nuclear weapons on people, animals, the environment, and various types of structures are urged to consult the exhaustive work by Glasstone and Dolan [1], which is the source of much of the material in this chapter.

  6.1 Brightness and thermal radiation

  If you have ever experienced a sunburn on a hot, clear summer day, you will have some sense of what weapons effects researchers refer to as ‘flash burns’.

  Shortly before the Trinity test, Los Alamos physicists Hans Bethe and Robert Christy prepared a document titled Memorandum on the Immediate After Effects of the Gadget [2]. In this report, they estimated that at about 1/100 of a second after the explosion, the surrounding air would be transformed into fireball of radius about 400 feet (∼125 m) and temperature ∼15 000°. I take the latter to be in Kelvins, although it makes little difference what units they intended. We can use these numbers to estimate the brightness of the fireball for an observer at a specified distance.

  Suppose that an observer is located at a distance of 2 km from the fireball, or a little more than a mile. The apparent size of the fireball against the sky can be estimated with simple trigonometry. The angular diameter of the fireball would be about (250 m/2000 m) = 0.125 radians, or about 7.2 degrees. The angular diameter of the Sun is about 0.5 degrees, so the fireball would appear about 14 times wider and cover an area of sky 142 ∼ 200 times that of the Sun!

  To estimate brightness, we can use the Stefan–Boltzmann law, which says that the total radiant power output of an object of radius R and absolute temperature T is proportional to R2T4. ‘Brightness’ is the power received per unit area by an observer, which for somebody located at distance d is the power divided by the area of the sphere over which the energy has spread, 4π d2. The brightnesses of the fireball (F) and Sun (S) will then compare as

  The Sun has RS = 6.96 × 108 m, effective surface temperature TS ∼ 5770 K, and is about 1.5 × 1011 m from the Earth. For the 125 m radius fireball with TF = 15 000 K observed from a distance of 2000 m,

  This comparison is approximate as not all of the power is radiated in the visible part of the spectrum, but since the temperatures are not too dissimilar the relevant coefficients would at least somewhat cancel when computing the brightness ratio. In any case, this result testifies to the staggering brightness of what is now regarded as a modest-energy nuclear weapon. It has been computed that the Trinity fireball would have momentarily appeared at least thirty times brighter than Venus for an observer located on the Moon, and would have been visible from Mars. Warning: even if you are protected from the shock wave and fallout created by a nuclear explosion, do not look at it: the focusing effect of your eyes can lead to serious retinal burns even miles away.

  The rate of delivery of radiant energy from a nuclear explosion is so prompt that combustible materials such as paper, wood, and clothing will be charred or burst into flame to great distances. Such materials can be ignited by the prompt delivery of 10 physical calories of radiant energy per square centimeter (1 physical calorie = 4.186 Joules); a 20 kt explosion delivers this much energy to a radius of about 6000 feet. At Trinity, some fir timbers were slightly scorched to this distance; such charring requires a temperature of about 400 °C. For human beings, moderate burns to unprotected skin can be produced by deposit of about 3 calories per square centimeter. For a 20 kt explosion, the radius for this effect is about 10 000 feet; at Nagasaki, skin burns were reported to 14 000 feet.

  The effects of such flash burns on humans depends on unpredictable factors such as atmospheric conditions, exposure duration, color of clothing (reflective or absorptive), and skin pigmentation. The resulting burns are classified as first, second, or third degree
. A bad sunburn is an example of a first-degree burn; a prompt exposure of 2–3 cal cm−2 will cause such burns for most people, although you will likely recover without any permanent scarring. Second-degree burns (∼4–5 cal cm−2) will develop scabs, but normally heal in a week or two unless an infection sets in. Third-degree burns (>∼6 cal cm−2) are the most damaging: burnt areas are so charred that they cannot transmit pain impulses, and so pain is felt only from surrounding areas. With such burns, skin grafts will be necessary to prevent scarring. To put these numbers in perspective, some 10–15 cal cm−2 are required to char trees; clothing and upholstery will typically ignite on exposure to 20–25 cal cm−2.

  Bearing in mind the above contingencies, there is available an approximate expression for thermal exposure. As with the energy released in a nuclear reaction, the symbol used for this is Q, and the formula is [1]

  Here Y is the weapon yield in kilotons, and d the distance of the observer from the fireball in miles. The factor τ is known as the ‘transmittance’, and is a measure of the attenuating effects of the atmosphere. (This τ has nothing to do with the travel time between fissions in the analysis of criticality in chapter 3.) For explosions within a few miles of the Earth’s surface and distances within a few miles of the detonation, a reasonable value is τ ∼ 0.7.

  For a 20 kt bomb at R = 2 miles and τ ∼ 0.7, equation (6.3) gives Q ∼ 4.2 cal cm−2, enough to cause a second-degree burn. It has been estimated that some two-thirds of those who died in the first day after the bombing of Hiroshima were badly burned. A 400 kiloton bomb at two miles will give Q ∼ 85 cal cm−2; you will literally be burnt alive.

  Exercise

  Verify the claim that a 20 kt bomb will give Q ∼ 10 cal cm−2 at a distance of 6000 feet. How far would you have to be from a 50 kt bomb to receive a 5 cal cm−2 second-degree burn exposure?

  Answer

  About 2.9 miles.

  6.2 Shock wave

  Most of the physical destruction caused by nuclear weapons is due to a high-pressure shock wave that propagates outward from the fireball. Normal atmospheric pressure is 14.7 pounds per square inch (psi). Weapons effects are usually stated in terms of the overpressure created, that is, the number of psi generated in excess of this ambient value. Seemingly small overpressures can have devastating effects. An overpressure of 1 psi is sufficient to break ordinary glass windows. Wood frame homes are destroyed under 5 psi overpressures, which is also about the threshold for human eardrum rupture. Eight to ten psi overpressure is sufficient to destroy brick houses and collapse factories and commercial buildings. Multistory buildings will sustain moderate damage at 6–7 psi overpressure, and will be demolished at 20 psi. Even if you are in no danger of being trapped within a collapsing structure, you are not necessarily safe: the threshold for human death from compressive effects sets in at about 40 psi.

  As with brightness and thermal effects, blast effects are influenced by meteorological conditions, terrain, surrounding structures, and particularly the height of the explosion above the ground. Here I will consider what weapons designers refer to as an ‘optimum height air burst’, an above-ground explosion that is at an altitude such that the resulting fireball just reaches the ground when it is at its brightest. In their book, Glasstone and Dolan include a slide-rule like device for estimating pressure effects for unshielded locations at a given ‘slant distance’ of d miles to the explosion for a weapon yield Y kilotons (figure 6.1). Numbers generated by this device can be used to develop a simple approximate formula for the maximum overpressure experienced, in psi:

  Consider the Hiroshima Little Boy bomb, which is estimated to have yielded ∼13 kt and was detonated at an altitude of ∼1900 feet. A point on the ground one mile from ground zero would have a slant range of ∼1.06 miles, for which equation (6.4) predicts Pmax ∼ 4.6 psi. Indeed, essentially everything was destroyed out to this distance except for some heavily reinforced concrete buildings. At a slant range of 2 miles from a Fat Man-like 20 kt yield, Pmax ∼ 2.2 psi: your house will be damaged, but you will likely survive if you can avoid flying debris, fallout, and thermal burns. Nowadays, yields of several hundred kilotons are now not uncommon; at two miles, a 400 kiloton yield will give an overpressure of nearly 10 psi, ensuring widespread devastation.

  Glasstone and Dolan offer the following approximate expression for the maximum radius of a fireball, in feet, for a weapon of yield Y kilotons:

  This expression is used in the following exercise.

  Exercise

  In a nuclear exchange with an adversary, you wish to subject an area within a ground distance s = 2 miles (figure 6.1) to an overpressure of at least 5 psi. Your policy is to detonate your weapons at a ‘burst height’ h equal to twice the fireball radius given by equation (6.5). What minimum weapon yield will you require to accomplish your mission?

  Figure 6.1. Geometry of burst height, slant distance, and ground distance.

  Answer

  about 110 kt.

  6.3 Neutron scattering

  Like any high-energy particle, neutrons cause biological damage by ionizing and dissociating atoms and molecules in the body of an organism exposed to them. Charged particles like alphas and betas are stopped fairly readily in air as they lose energy by causing ionizations, but neutrons are uncharged and so this mechanism is much less effective for them; their only real energy loss results from collisions with nuclei of molecules in the air. In this section I estimate the distance to which neutrons might be expected to be dangerous. I break here with the pattern of preceding chapters and offer a brief derivation of this effect to illustrate how some basic principles of mechanics can be used to estimate a weapons effect.

  The distance that a neutron will propagate through a medium depends on two factors: how many collisions are needed to reduce its kinetic energy to a level considered harmless, and how far it travels on average between collisions, that is, its mean free path. Look at figure 6.2. Suppose that a neutron with some initial speed v0 strikes an initially stationary nucleus, and scatters at some angle a; the nucleus recoils at angle b. Let the neutron have speed vn after the collision, and the nucleus speed V.

  Figure 6.2. Neutron–nucleus collision. The struck nucleus is initially stationary.

  If the nucleus is of mass number A, we can take the neutron to be of mass 1. If the collision is elastic, that is, if both momentum and kinetic energy are conserved, it is possible to solve for the speed of the neutron after the collision:

  If you are familiar with Compton scattering, the derivation of this expression is very similar to that for the change in wavelength of the scattered photon in that phenomenon.

  If the neutron barely grazes the nucleus and continues on its way, that is, if a ∼ 0, then ∼ 1, and the neutron loses essentially no kinetic energy. The other extreme is a head-on collision and backwards recoil, that is, a = 180°. In this case = (A − 1)/(A + 1), and the neutron loses some kinetic energy. Most collisions will be between these extremes. For sake of simplicity I will assume that the average neutron post-collision speed is given by the average of these extremes, ∼ A/(A + 1). The ratio of post-collision to pre-collision kinetic energies will then on average be ∼ . If a neutron suffers ω successive collisions, then it follows that . Now, the goal here is to get an expression for the number of collisions required to reduce the neutron’s kinetic energy to some specified value. We could just solve for ω, but a more compact result can be obtained by invoking some convenient approximations. First, take the logarithm of this result:

  The logarithm on the right side can be simplified as follows. Extract a factor of A from both numerator and denominator to give

  where a binomial expansion was performed in the second-last step, and the approximation ln (1 + x) ∼ x was applied in the last step. (This approximation is valid for −1 ⩽ x ⩽ 1.) With this, we have the desired result:

  Consider fission-liberated neutrons of Einitial ∼ 2 MeV traveling through air. Air is mostly (∼80%) diatomic nitrogen, which has an ato
mic weight of 28. At room-temperature conditions, particle kinetic energies are typically ∼1/25 eV (from 3kT/2), so I will assume that it is desired to slow neutrons to this energy in order to consider them harmless. Hence

  To estimate the distance traveled between collisions, we can invoke the mean-free-path concept that was used in the analysis of critical mass and tampering in chapter 3:

  For atomic nitrogen, the fission-spectrum-averaged scattering cross-section for neutrons is about 1.8 barns; I double this to account for molecular nitrogen. Air has a density of about 1.2 kg m−3, of which I assume that N2 comprises about 1 kg m−3. In combination with the molecular weight of 28 g mol−1, this gives n ∼ 2.15 × 1025 molecules m−3, and a mean free path of λ ∼ 130 m.

  Neutrons will scatter from nitrogen molecules in all directions; to estimate how far they diffuse from the explosion, we can model their flight as a random-walk process. For ω collisions each of step length λ, the mathematics of random walks shows that average distance neutrons reach from their starting point is . For the numbers above, d ∼ 2000 m, or a little over a mile. If you are in the vicinity of a nuclear explosion, your more immediate concerns are likely to be heat and blast, but the ‘invisible’ effects of neutron irradiation are not to be discounted.

 

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