The Los Alamos Primer — Section III

“In a great number of cases, I have seen Dr. Oppenheimer act — I understand that Dr. Oppenheimer acted — in a way which was for me was exceedingly hard to understand. I thoroughly disagreed with him in numerous issues and his actions frankly appeared to me confused and complicated. To this extent I feel that I would like to see the vital interests of this country in hands which I understand better, and therefore trust more. In this very limited sense I would like to express a feeling that I would feel personally more secure if public matters would rest in other hands.”

Edward Teller — Oppenheimer’s security hearing in 1954

Oppenheimer’s former colleague, physicist Edward Teller, testified on behalf of the government at Oppenheimer’s security hearing in 1954.

The ‘Los Alamos Primer’ is perhaps the most important single document of the 20th Century. Yet, it is a rather simple document that could have been written by many others “elsewhere”, and possibly several years earlier. Thus, the History of this world could have been drastically different. In this post, I  discus the  Section III of the  document: Fast Neutron Chain Reaction. Follow us on Twitter: @INTEL_TODAY

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Here is Section 3 of the document followed by my analysis and comments.

THE LOS ALAMOS PRIMER

— — —

3. Fast Neutron Chain Reaction

Release of this energy in a large scale way is a possibility because of the fact that in each fission process, which requires a neutron to produce it, two neutrons are released.

Consider a very great mass of active material, so great that no neutrons are lost through the surface and assume the material so pure that no neutrons are lost in other ways than by fission. One neutron released in the mass would become 2 after the first fission, each of these would produce 2 after they each had produced fission so in the nth generation of neutrons there would be 2n neutrons available.

Since in 1 kg. of 25 there are 5 . 10^25 nuclei it would require about n = 80 generations (2^80 = 5 . 10^25) to fissile the whole kilogram.

(Page 2)

While this is going on the energy release is making the material very hot, developing great pressure and hence tending to cause an explosion.

In an actual finite setup some neutrons are lost by diffusion out through the surface. There will be therefore a certain size of say a sphere for which the surface losses of neutrons are just sufficient to stop the chain reaction.

This radius depends on the density. As the reaction proceeds the material tends to expand, increasing the required minimum size faster than the actual size increases.

The whole question of whether an effective explosion is made depends on whether the reaction is stopped by this tendency before an appreciable fraction of the active material has fissiled.

Note that the energy released per fission is large compared to the total binding energy of the electrons in any atom. In consequence even if but 1/2% of the available energy is released the material is very highly ionized and the temperature is raised to the order of 40 . 10^6 degrees. If 1% is released the mean speed of the nuclear particles is of the order of 10^8 cm/sec. Expansion of a few centimeters will stop the reaction, so the whole reaction must occur in about 5 . 10^8 sec otherwise the material will have blown out enough to stop it.

Now the speed of a 1 MEV neutron is about 1.4 . 10^9 cm/sec and the mean free path between fissions is about 13 cm so the mean time between fissions is about 10 ^ -8 sec.

Since only the last few generations will release enough energy to produce much expansion, it is just possible for the reaction to occur to an interesting extent before it is stopped by the spreading of the active material.

Slow neutrons cannot play an essential role in an explosion process since they require about a microsecond to be slowed down in hydrogenic materials and the explosion is all over before they are slowed down.

Analysis and discussion

Serber took ν = 2 for U-235. The accepted value is higher: ν = 2.637.

Serber mistakenly gives the number of nuclei in 1 kg of U-235 as 5 × 10^25; the correct answer is: N = 2.56 × 10^24.

“… it would require about n = 80 generations (2^80 = 5 . 10^25) to fissile the whole kilogram.”

That is not correct. 2^80 = 1.2 x 10^24! The correct answer is n = 81 as 2^81 = 2.4 x 10^24. Notice that n decreases rapidly with increasing values of ν. With ν = 2.637, n is equal to about 58 generations.

“The whole question of whether an effective explosion is made depends on whether the reaction is stopped by this tendency before an appreciable fraction of the active material has fissiled.”

As the core expands quickly, it reaches a ‘radius’ beyond which criticality is no longer maintained and the chain reaction stops. How much energy has been generated before the device reaches ‘second criticality’ defines the effiency of the bomb.

“Since only the last few generations will release enough energy to produce much expansion, it is just possible for the reaction to occur to an interesting extent before it is stopped by the spreading of the active material.”

Notice the words: “just possible”.  In the next post, I will discuss the Bethe-Feynman formula which provides an important clue to the solution of this problem.

Hans Bethe: How I met Feynman at Los Alamos

REFERENCES

Revisiting The Los Alamos Primer — Physics Today

The Los Alamos Primer — Original pdf

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The Los Alamos Primer — Section III

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