Small yield of tactical nuclear weapons

  • Thread starter Thread starter Thecla
  • Start date Start date
  • Tags Tags
    Nuclear Yield
AI Thread Summary
Uranium fission bombs typically have an explosive yield of about 10,000 tons of TNT, and creating a bomb with a yield of 5 megatons is not feasible due to the critical mass requirements of U-235. Tactical nuclear weapons can achieve low yields, such as 10 tons of TNT, through advanced design techniques rather than using pure uranium. Older uranium-based devices have largely been decommissioned in favor of more modern designs that utilize boosting methods, such as deuterium and tritium (D+T), to enhance energy release. The specifics of nuclear explosive design are not publicly disclosed, emphasizing the sensitive nature of this information. Overall, the discussion highlights the complexities and advancements in nuclear weapon technology.
Thecla
Messages
136
Reaction score
10
TL;DR Summary
How are the small yields of tactical nuclear weapons achieved?
Uranium fission bombs have explosive power ofabout 10 000 tons of TNT. I understand that you can't make a uranium bomb with explosive power of 5 megatons of TNT brcause of the critical mass of U-235. Similarly you can't make an atomic fission weapon with explosive power of 10 tons of TNT because the mass of U-235 would be below the critical mass. So how do tactical nuclear weapons achieve such low yields?
 
Engineering news on Phys.org
It would be a waste of material for 10 ton TNT U-based nuclear device. Older U devices have been decommissioned in favor more modern devices. To obtain energy releases in excess of 10 kT TNT-eq, devices are boosted with D+T. Design of nuclear explosives are not for public discussion.
 
  • Like
Likes topsquark
Astronuc said:
Design of nuclear explosives are not for public discussion.
And with that, this thread is closed.
 
  • Like
Likes topsquark
Hello, I'm currently trying to compare theoretical results with an MCNP simulation. I'm using two discrete sets of data, intensity (probability) and linear attenuation coefficient, both functions of energy, to produce an attenuated energy spectrum after x-rays have passed through a thin layer of lead. I've been running through the calculations and I'm getting a higher average attenuated energy (~74 keV) than initial average energy (~33 keV). My guess is I'm doing something wrong somewhere...

Similar threads

Replies
66
Views
20K
Replies
9
Views
3K
Replies
4
Views
2K
Replies
4
Views
2K
Replies
17
Views
3K
Replies
4
Views
2K
Replies
5
Views
3K
Replies
13
Views
6K
Back
Top