In anuclear weapon, atamper is an optional layer of dense material surrounding thefissile material. It is used innuclear weapon design to reduce thecritical mass and to delay the expansion of the reacting material through itsinertia, which delays the thermal expansion of the fissioning fuel mass, keeping it supercritical longer. Often the same layer serves both as tamper and asneutron reflector. The weapon disintegrates as the reaction proceeds, and this stops the reaction, so the use of a tamper makes for a longer-lasting, more energetic and more efficient explosion. Theyield can be further enhanced using a fissionable tamper.

The first nuclear weapons used heavynatural uranium ortungsten carbide tampers, but a heavy tamper necessitates a largerhigh-explosive implosion system and makes the entire device larger and heavier. The primary stage of a modernthermonuclear weapon may instead use a lightweightberyllium reflector, which is also transparent toX-rays whenionized, allowing the primary's energy output to escape quickly to be used in compressing the secondary stage. More exotic tamper materials such asgold are used for special purposes like emitting large amounts of X-rays or altering the amount ofnuclear fallout.

While the effect of a tamper is to increase efficiency, both by reflectingneutrons and by delaying the expansion of the bomb, the effect on the critical mass is not as great. The reason for this is that the process of reflection is time-consuming. By the time reflected neutrons return to thecore, several generations of thechain reaction have passed, meaning the contribution from the older generation is a tiny fraction of the neutron population.

InAtomic Energy for Military Purposes (1945), physicistHenry DeWolf Smyth describes the function of a tamper innuclear weapon design as similar to theneutron reflector used in anuclear reactor:

A similar envelope can be used to reduce thecritical size of the bomb, but here the envelope has an additional role: its veryinertia delays the expansion of the reacting material. For this reason such an envelope is often called a tamper. Use of a tamper clearly makes for a longer lasting, more energetic and more efficient explosion.^[1]

The concept of surrounding thecore of anuclear weapon with a tamper was introduced byRobert Serber in hisLos Alamos Primer, a series of lectures given in April 1943 as part of theManhattan Project, which built the first nuclear weapons. He noted that sinceinertia was the key, the densest materials were preferable, and he identifiedgold,rhenium,tungsten anduranium as the best candidates. He believed they also had goodneutron-reflecting properties, although he cautioned that a great deal more work needed to be done in this area. Using elementarydiffusion theory, he predicted that thecritical mass of a nuclear weapon with a tamper would be one-eighth that of an identical but untamped weapon. He added that in practice this would only be about a quarter instead of an eighth.^[2][3]

Serber noted that the neutron reflection property was not as good as it might first seem, because the neutrons returning from collisions in the tamper would take time to do so. He estimated that for a uranium tamper they might take about 10⁻⁷ seconds. By the time reflected neutrons return to the core, several generations of thechain reaction would have passed, meaning the contribution from the older generation is a tiny fraction of the neutron population. The returning neutrons would also be slowed by the collision. It followed that 15% morefissile material was required to get the same energy release with a gold tamper compared to a uranium one, despite the fact that the critical masses differed by 50%.^[4] At the time, the critical masses of uranium (and more particularlyplutonium) were not precisely known. It was thought that uranium with a uranium tamper might be about 25 kg, while that of plutonium would be about 5 kg.^[3]

TheLittle Boy uranium bomb used in theatomic bombing of Hiroshima had atungsten carbide tamper. This was important not just for neutron reflection but also for its strength in preventing the projectile from blowing through the target.^[5] The tamper had a radius of 17.5 centimetres (6.9 in) and a thickness of 11.3 centimetres (4.4 in), for a mass of 317 kilograms (699 lb). This was about 3.5 times the mass of the fissile material used. Tungsten carbide has a high density and a low neutron absorbencycross section. Despite being available in adequate quantity during the Manhattan Project,depleted uranium was not used because it has a relatively high rate ofspontaneous fission of about 675 per kg per second; a 300 kg depleted uranium tamper would therefore have an unacceptable chance of initiating apredetonation.^[6] Tungsten carbide was commonly used inuranium-233gun-type nuclear weapons used with artillery pieces for the same reason.^[7][8]

There are advantages to using a fissionable tamper to increase the yield.Uranium-238 will fission when struck by a neutron with 1.6megaelectronvolts (0.26 pJ), and about half the neutrons produced by the fission ofuranium-235 will exceed this threshold. However, afast neutron striking a uranium-238 nucleus is eight times as likely to be inelastically scattered as to produce a fission, and when it does so, it is slowed to the point below the fission threshold of uranium-238.^[9] In theFat Man type used in theTrinity test and at Nagasaki, the tamper consisted of 7.0-centimetre (2.75 in) shells of natural uranium andaluminium.^[5][10] It is estimated that up to 30% of theyield came from fission of thenatural uranium tamper.^[11][12] An estimated 14.5 tonnes of TNT (61 GJ) of the 21 kilotonnes of TNT (88 TJ) yield was contributed by thephotofission of the tamper.^[13]

In aboosted fission weapon or athermonuclear weapon, the 14.1-megaelectronvolt (2.26 pJ) neutrons produced by adeuterium-tritium reaction can remain sufficiently energetic to fission uranium-238 even after three collisions with deuterium, but the 2.45-megaelectronvolt (0.393 pJ) ones produced by deuterium-deuterium fusion no longer have sufficient energy after even a single collision. A uranium-235 tamper will fission even with slow neutrons. A highlyenriched uranium tamper is therefore more efficient than a depleted uranium one, and a smaller tamper can be used to achieve the same yield. The use of enriched uranium tampers therefore became more common once enriched uranium became more plentiful.

An important development after World War II was the lightweightberyllium tamper. In a boosted device the thermonuclear reactions greatly increase the production of neutrons, which makes the inertial property of tampers less important. Beryllium has a low slow neutron absorbency cross section but a very high scattering cross section. When struck by high energy neutrons produced by fission reactions, beryllium emits neutrons. With a 10-centimeter (4 in) beryllium reflector, the critical mass of highly enriched uranium is 14.1 kg, compared with 52.5 kg in an untamped sphere. A beryllium tamper also minimizes the loss of X-rays, which is important for a thermonuclear primary which uses its X-rays to compress the secondary stage.^[14]

The beryllium tamper had been considered by the Manhattan Project, but beryllium was in short supply, and experiments with a beryllium tamper did not commence until after the war. PhysicistLouis Slotin was killed in May 1946 in acriticality accident involving one. A device with a beryllium tamper was successfully tested in theOperation Tumbler–Snapper How shot on 5 June 1952, and since then beryllium has been widely used as a tamper in thermonuclear primaries.^[14] The secondary's tamper (or "pusher") functions to reflect neutrons, confine the fusion fuel with its inertial mass, and enhance the yield with its fissions produced by neutrons emitted from the thermonuclear reactions. It also helps drive the radiation implosion and prevent the loss of thermal energy. For this reason, the heavy tamper is still preferred.^[15]

Thorium can also be used as a fissionable tamper. It has an atomic weight nearly as high as uranium and a lower propensity to fission, which means that the tamper has to be much thicker.^[15] It is possible that a state seeking to develop nuclear weapons capability might addreactor-grade plutonium to a natural uranium tamper. This would cause problems with neutron emissions from the plutonium, but it might be possible to overcome this with a layer ofboron-10,^[15] which has a high neutron cross section for the absorption of the slow neutrons that fission uranium-235 andplutonium-239, but a low cross-section for the absorption of the fast neutrons that fission uranium-238. It was used in thermonuclear weapons to protect the plutonium spark plug from stray neutrons emitted by the uranium-238 tamper.^[16] In the Fat Man type the natural uranium tamper was coated withboron.^[17]

Non-fissionable materials can be used as tampers. Sometimes these were substituted for fissionable ones innuclear tests where a high yield was unnecessary.^[18] The most commonly used non-fissionable tamper material islead, which is both widely available and cheap. British designs often used a lead-bismuth alloy. Bismuth has the highest atomic number of any non-fissionable tamper material. The use of lead and bismuth reducesnuclear fallout, as neither produces isotopes that emit significant amounts ofgamma radiation when irradiated with neutrons.^[15]

TheW71 warhead used in theLIM-49 Spartananti-ballistic missile had a gold tamper around its secondary to maximize its output of X-rays, which it used to incapacitate incoming nuclear warheads.^[15][19] The irradiation of gold-197 produces gold-198, which has ahalf-life of 2.697 days and emits 0.412-megaelectronvolt (0.0660 pJ) gamma rays and 0.96-megaelectronvolt (0.154 pJ)beta particles. It therefore produces short-lived but intense radiation, which may have battlefield uses, although this was not its purpose in the W71. Another element evaluated by the US for such a purpose wastantalum. Natural tantalum is almost entirely tantalum-181, which when irradiated with neutrons become tantalum-182, a beta and gamma ray emitter with a half-life of 115 days.

In the theoreticalcobalt bomb,^[15]cobalt is poor prospect for a tamper because it is relatively light andionizes at 9.9 kiloelectronvolts (1.59 fJ). Natural cobalt is entirely cobalt-59, which becomescobalt-60 when irradiated with neutrons. With a half-life of 5.26 years, this could produce long-lasting radioactive contamination.^[15] The BritishTadje nuclear test atMaralinga used cobalt pellets as a "tracer" for determining yield.^[20] This fuelled rumours that Britain had been developing a cobalt bomb.^[21]

The diffusion equation for the number of neutrons within a bomb core is given by:^[22]

${\displaystyle \frac{\mathrm{\partial }N}{\mathrm{\partial }t}=\frac{v_n}{{\lambda }_f^{core}}(\nu -1)N+\frac{{\lambda }_t^{core}{v}_n}{3}({\mathrm{\nabla }}^{2}N)}$

where${\displaystyle N}$ is the number density of neutrons,${\displaystyle v_n}$ is the average neutron velocity,${\displaystyle \nu }$ is the number of secondary neutrons produced per fission,${\displaystyle {\lambda }_f^{core}}$ is the fissionmean free path and${\displaystyle {\lambda }_t^{core}}$ is transport mean free path for neutrons in the core.

${\displaystyle N}$ doesn't depend on the direction, so we can use this form of theLaplace operator in spherical coordinates:

${\displaystyle {\mathrm{\nabla }}^{2}N=\frac{1}{r^2}\frac{\mathrm{\partial }}{\mathrm{\partial }r}(r^2\frac{\mathrm{\partial }N}{\mathrm{\partial }r})}$

Solving theseparable partial differential equation gives us:^[23]

${\displaystyle N_{core}(r,t)=N_0{e}^{(\alpha /\tau )t}[\frac{sin(r/d_{core})}{r}]}$

where

${\displaystyle \tau ={\lambda }_f^{core}/v_n}$

and

${\displaystyle d_{core}=\sqrt{\frac{{\lambda }_f^{core}{\lambda }_t^{core}}{3(-\alpha +\nu -1)}}}$

For the tamper, the first term in the first equation relating to the production of neutrons can be disregarded, leaving:

${\displaystyle \frac{\mathrm{\partial }N}{\mathrm{\partial }t}=\frac{{\lambda }_t^{core}{v}_n}{3}({\mathrm{\nabla }}^{2}N)}$

Set the separation constant as${\displaystyle \delta /\tau }$. If${\displaystyle \delta =0}$ (meaning that the neutron density in the tamper is constant) the solution becomes:

${\displaystyle N_{tamper}=\frac{A}{r}+B}$

Where${\displaystyle A}$ and${\displaystyle B}$ areconstants of integration.

If${\displaystyle \delta >0}$ (meaning that the neutron density in the tamper is growing) the solution becomes:^[24]

${\displaystyle N_{tamper}=e^{(\delta /\tau )t}[A\frac{e^{r/d_{tamper}}}{r}+B\frac{e^{-r/d_{tamper}}}{r}]}$

where

${\displaystyle d_{tamper}=\sqrt{\frac{{\lambda }_f^{core}{\lambda }_t^{core}}{3}}}$

Serber noted that at the boundary between the core and the tamper, the diffusion stream of neutrons must be continuous,^[2] so if the core has radius${\displaystyle R_{core}}$ then:^[24]

${\displaystyle N_{core}(R_{core})=N_{tamper}(R_{core})}$

If the neutron velocity in the core and the tamper is the same, then${\displaystyle \alpha =\delta }$ and:^[24]

${\displaystyle {\lambda }_t^{core}{(\frac{\mathrm{\partial }{N}_{core}}{\mathrm{\partial }r})}_{R_{core}}={\lambda }_t^{tamper}(\frac{\mathrm{\partial }{N}_{tamper}}{\mathrm{\partial }r}{)}_{R_{core}}}$

Otherwise each side would have to be multiplied by the relevant neutron velocity. Also:^[24]

${\displaystyle N_{tamper}(R_{tamper})=-\frac{2}{3}{\lambda }_t^{tamper}(\frac{\mathrm{\partial }{N}_{tamper}}{\mathrm{\partial }r}{)}_{R_{core}}}$

For the case where${\displaystyle \alpha =\delta =0}$:

${\displaystyle [1+\frac{2R_{tamper}^{threshold}{\lambda }_t^{tamper}}{3R_{tamper}^2}-\frac{R_{tamper}^{threshold}}{R_{tamper}}][(\frac{R_{tamper}^{threshold}}{d_{core}})cot(\frac{R_{tamper}^{threshold}}{d_{core}})-1]+\frac{{\lambda }_t^{tamper}}{{\lambda }_t^{core}}=0}$

If the tamper is really thick, ie${\displaystyle R_{tamper}\gg R_{tamper}^{threshold}}$ this can be approximated as:

${\displaystyle (\frac{R_{tamper}^{threshold}}{d_{core}})cot(\frac{R_{tamper}^{threshold}}{d_{core}})=1-\frac{{\lambda }_t^{tamper}}{{\lambda }_t^{core}}}$

If the tamper (unrealistically) is a vacuum, then the neutron scattering cross section would be zero and${\displaystyle {\lambda }_t^{tamper}=\mathrm{\infty }}$. The equation becomes:

${\displaystyle (\frac{R_{tamper}^{threshold}}{d_{core}})cot(\frac{R_{tamper}^{threshold}}{d_{core}})=-\mathrm{\infty }}$

which is satisfied by:

${\displaystyle (\frac{R_{tamper}^{threshold}}{d_{core}})=\pi }$

If the tamper is very thick and has neutron scattering properties similar to the core, ie:

${\displaystyle {\lambda }_t^{tamper}\sim {\lambda }_t^{core}}$

Then the equation becomes:

${\displaystyle (\frac{R_{tamper}^{threshold}}{d_{core}})cot(\frac{R_{tamper}^{threshold}}{d_{core}})=0}$

which is satisfied when:

${\displaystyle \frac{R_{tamper}^{threshold}}{d_{core}}=\pi /2}$

In this case, the critical radius is twice what it would be if no tamper were present. Since the volume is proportional to the cube of the radius, we reach Serber's conclusion that an eightfold reduction in the critical mass is theoretically possible.^[2][25]

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