Amount of Plutonium used in Trinity

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Homework Help Overview

The discussion revolves around estimating the amount of Plutonium used in the 1945 Trinity test by analyzing energy release through dimensional analysis and the equation E=mc^2, considering that only a fraction of the rest mass was converted to energy during fission.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to estimate energy released and mass of Plutonium using dimensional analysis and E=mc^2. Participants question the methodology and the assumptions made regarding proportionality constants and the interpretation of the energy distribution.

Discussion Status

Participants are actively engaging with the original poster's calculations, with some expressing uncertainty about the accuracy of the results and others suggesting that the approach may be fundamentally sound despite potential errors in constants or assumptions.

Contextual Notes

There is mention of a specific expected mass of around 6 kg, which is derived from external sources rather than explicit instructions from the instructor. The discussion also highlights uncertainties related to energy distribution and losses during the explosion.

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Amount of Plutonium used in "Trinity"

Hi,

Homework Statement



I am asked to find the amount of Plutonium used in the 1945 experiment, first by estimating the energy released based on dimensional analysis and a photo, then using E=mc^2 and the fact that merely 0.1% of the rest mass was released during the fission.


Homework Equations





The Attempt at a Solution



I first arrived at 1.0023*10^14 Joules (which is pretty close to the official results (8.78*10^13 Joules), then the mass would have to be 1000*E/(c^2). 1000 due to the 0.1% actually released. Which yielded 1.113 kg, but that's way off! The correct answer has to be around 6 kg. What am I doing wrong?
 
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How did you first arrive at 1.0032*1014J?
 


From dimensional analysis, E is proportional to rho*R^5/t^2. Rho is around 1.298 kg/m^3, R (based on the photo) is 137 m, and t=0.025 sec. This yields E=1.0023*10^14 Joules.
 


It's rather surprising no one has yet commented on the solution and the potential error.
 


probably because you didn't explain your work very much, so your question is confusing. What photo are you talking about? What work did you do to get E is proportional to ρR5/t2?
 


As I said, sheer dimensional analysis, i.e. equating units. Hence, I wasn't reluctant to explain anything, there is simply not much to explain. The photo may be found online and the link was given to us by the instructor:
http://www.flickr.com/photos/62592750@N05/5699376671/
 


Sorry if I implied that I thought you were "reluctant" :)

maybe it's just to do with whatever constant of proportionality might be there? I mean, you're off by less than an order of magnitude, and that's pretty good for rough "back of the envelope" calculations. Was your instructor very specific about wanting 6kg?
 


Not at all. I found the "official" result online. Is there anything amiss with my calculation? Can you point out anything specific? I mean, aside a possible constant of proportionality.
 


I'm not exactly sure, sorry. My feeling is that you are close enough for this kind of problem. The energy is probably not distributed evenly inside of the fireball even after just 0.025s, so that might be a source of uncertainty. These test bombs were also detonated while suspended some number of meters off the ground, right? A good portion of the energy is also probably lost into the ground.

I think that you approached the problem correctly and went through it with the right ideas, and that's what really matters. Not so much having an answer that's off by a bit.
 
  • #10


Thank you!
 

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