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Q: nuclear energy of the human body ( Answered 4 out of 5 stars,   3 Comments )
Subject: nuclear energy of the human body
Category: Science > Physics
Asked by: dudester123-ga
List Price: $10.00
Posted: 08 Jun 2006 23:56 PDT
Expires: 08 Jul 2006 23:56 PDT
Question ID: 736619
I understand that a single nuclear fission reaction of one uranium
atom is enough to make a grain of sand jump. So, how much nuclear
energy is in the human body? How can I quantify it? I understand that
fission can only take place with uranium or plutonium. But is it
possible to quantify the amount of nuclear energy in a human body?
Subject: Re: nuclear energy of the human body
Answered By: hedgie-ga on 10 Jun 2006 00:19 PDT
Rated:4 out of 5 stars
Elements with mass number larger then Iron may produce energy by fission

 --------- comment on comments--\\
 This was  correctly noted yonr.  But then he switched to C-14, which
radioactive, not fissionable.
Confusion was futher compouded by qed, which is not unususal, who said

" all atoms conrain nucleons" 

That is true, but  that does not make them 'nuclear materials' .

 Nuclear Materials are those used in the nuclear industry, not those
having some nucleons in thehm
Special Nuclear Material - consists of uranium-233 or uranium-235,
enriched uranium, or plutonium


Let's continue the answer with respect to energy which can be released by fission.

 Elements with mass number larger then Iron may produce energy by fission

Those elements are shown here  (click on atomic mass, , Iron Fe is abour  55 )

binding energy curve is shown here:

Sample calculation: energy released in nuclear fission

It has Sample calculation: energy released in nuclear fission at the
bottom of the page

You can expect some 10 MeV in sub-uranium fission, so multiply that bu
number of atoms
in human body which has mass number above 55.

Elements in human body are listed here:

   Heavy elelents are present only as traces (heavy metals are toxic)

                       mass number
traces	Copper		63
traces	Zinc		65
traces	Selenium	78	
traces	Molybdenum      95

You can ask for additional details here:

Clarification of Answer by hedgie-ga on 10 Jun 2006 01:15 PDT
To be more exact, qed100 did not say exactly
" all atoms conrain nucleons"

(He does not knows much about inovative spelling, I often use :)

but 'Every single atom has nuclear matter within it .."

However, binding energy of all nucleons, or  energy equivalent of all mass
etc etc are not same as   energy which can be released by fission, which is
(very aproximately) as follows : 

  Let take as 'typical atom' in the human body
  which is fissionable to have mass number 60
 6.0221415E23 atoms is one mole 
 one mole of such atoms weight about 60 g

Based on table of trace elements:

 Roughly, human body has about 1mg of trace elements 
 with atomic weight around 60


1 mg  of mass number 60  element has  6E23/60E3 = 1e19 atoms  

one such atom gives out about  10 MeV in a fission reaction


One human body used as a fuel in a fission reactor would yield about

 1E19 * 10 MeV 

enter into  Google search engine

 1.E19*10 MeV in Joules

and you get

 	1.E19 * 10 megaelectron volts = 16 021 764.6 joules

 1.E19*10 MeV in BTU

will give you

 	1.E19 * 10 megaelectron volts = 15 185.7028 BTU


Proper unit for energy are Joules. 

 For comparison,
 One 100W bulb uses 100 J every second, so you would get some

16 021 seconds = 4.45027778 hours

of ilumination.

Clarification of Answer by hedgie-ga on 26 Jun 2006 09:29 PDT
I am sorry if I did that.
It is not easy the gauge from 5 line question
what customer knows - so I prefer to err on assuming less.
dudester123-ga rated this answer:4 out of 5 stars
Please, don't worry about insulting me by talking down to me....

Subject: Re: nuclear energy of the human body
From: yonr-ga on 09 Jun 2006 09:12 PDT
The only nuclear matter that I'm aware is present in the human body in
any noteworthy quantity is C-14.

One atom of C-14 produces 0.156 mega-electron volts
( This is equivalent
to 1.60 * 10^-13 J, or 2.27*10^-11 lb*in. So one C-14 atom can lift
2.27*10^-11 lbs 1 in, or 1 lb 2.27*10^-11 ins, and the proportional

Now the question is how much C-14 is there in the human body...

There are 95 micrograms (10^-6 g) of C14 in the body.

Because C14's atomic mass is 14, multiply 95/1,000,000 by 14 to get
the number of moles of atoms. We have (95/1,000,000)(14) = 1.33*10^-3.
Since there are 6.0225  10^23 atoms in every mole, multiply this by
6.0225  10^23.

(1.33*10^-3)*(6.0225  10^23) = 8.01*10^20.

But this only gives the number of atoms. Now we need to multiply by
the work done by each atom:

(8.01*10^20 atoms)*(1.60 * 10^-13 J/atom) = 1.28*10^8 J.

So you can move 1.28*10^8 Newtons one meter. At Earth's surface, that
would be just about 1.28*10^7 kg, or 1.28*10^4 metric tons. So you can
lift about 1,280 tons one meter into the air with the nuclear energy
in your body. That's about 250 elephants! Or you could toss one
elephant 250 meters (762 feet). Or anything proportional.
Subject: Re: nuclear energy of the human body
From: qed100-ga on 09 Jun 2006 09:29 PDT

   What do you mean, the only "nuclear" matter in the human body is
C-14. Every single atom has nuclear matter within it: the nucleons
themselves. It's not an issue of the radioactivity of an element. In
anything more massive than straight hydrogen, there is energy
associated with holding the nucleons together. That energy is the
nuclear energy of the atom.

   As for dudester123's question, it is possible to quantify the
nuclear energy in the human body. To arrive at a very precise number
would, however, require a thorough breakdown of all the chemical
elements' abundances in the body (how many atoms of each element are
present), and then multiplying those numbers by the binding energies
of the various elements' nuclei (including by isotope).
Subject: Re: nuclear energy of the human body
From: yonr-ga on 09 Jun 2006 12:22 PDT
qed100, it appears that I misinterpreted the question. Still, I think
it's kind of amusing that the decay of the C14 in our bodies can lift

As for the nuclear matter in the human body... for all the elements up
to Iron and nickle it takes more energy to separate the nucleus than
you get from dividing it. Fission doesn't really pay off energywise
for most stable atoms. That's why fusion produces energy for the
combination of atoms up to Fe and Ni, and actually consumes energy
thereafter. The reverse can be said for fission.

If you care about the fusion of the elements that make up 96% of the
body, we're dealing with 10% of the body's mass in hydrogen, 63% in
oxygen and 23% in carbon. The other 4% is negligable. For a 60 kg
person (about 132 lbs) that would be 6 kg hydrgen, 38 kg oxygen and 14
kg carbon.

Carbon and oxygen don't really synthesize well, but if you want to
know how much energy can hypothetically be drawn from them in a
perfect system, I'll get there in a second.

Supposing we could conduct all the hydrogen perfectly (the
deuterium-tritium process), 5 atoms should produce 17 MeV. Again we do
the stoichiometry:

(15 kg)*(1000 g/kg)*(1 mol/1 g)*(6.0225*10^23 atoms/mol) = 9.03*10^27 atoms.

(9.03*10^27 atoms)*(17 MeV/5 atoms) = 3.07*10^28 MeV

(3.07*10^28 MeV)*(1.6 x 10^-13 J/MeV) = 4.91*10^15 J

This on earth's surface would lift about 4.91*10^14 kg-meters. That'll
launch an elephant 100,000,000 kilometers (although Earth's gravity
wouldn't really hold him at that point). This would be an opportune
time to correct my calculation on C14: I forgot to multiply by
1.6*10^-13 J/MeV. Alas, our radioactive carbon would only launch an
elephant 32 meters. If you find similar errors here (although I'm
checking more scropulously, because the first was done right as I was
exiting the house for an appointment), let me know. I'm curious to
know the answer to this question as well.

The most energy-efficient fusion of oxygen produces 16.5 MeV for every
two atoms of oxygen used.

(38 kg)*(1000 g/kg)*(1 mol/16 g)*(6.0225*10^23 atoms/mol) = 1.43*10^27 atoms

(1.43*10^27 atoms)*(16.5 Mev/2 atoms) = 1.18*10^28 MeV

(1.18*10^28 MeV)*(1.6 x 10^-13 J/MeV) = 1.89*10^15 J

On earth, that would do about 1.89*10^14 kg-meters of work, lifting
about 40% of an elephant 100,000,000 kilometers.

Assuming the most energy-efficient path (to Mg24), carbon would
release 14 MeV of energy for every two atoms
Stoichiometry again:

(14 kg)*(1000 g/kg)*(1 mol/12 g)*(6.0225*10^23 atoms/mol) = 7.03*10^26 atoms.

(7.03*10^26 atoms)*(14 MeV/2 atoms) = 4.92*10^27 MeV

(4.92*10^27 MeV)*(1.6 x 10^-13 J/MeV) = 7.87*10^14 J

At sea level, this would carry 7.87*10^13 kg-meters. This would lift
an elephant and a half about 10,000,000 kilometers.

So, for our grand totaly, we have 7.59*10^15 J, or about 7.59*10^14
kg-meters, or 7.59*10^8 ton-km at sea level. So we can throw an
elephant about 150,000,000 km (a little under 100,000,000 miles).

Let's see how much that is relative to our distance to the sun:

We are 8.3 light minutes away from the sun. Light travels 300,000 km/sec.
(300,000 km/sec)*(60 sec/1 min)*(8.3 min) = 1.49*10^8 km So we're
1.49*10^8 km away from the sun. That's practically identical to our
elephant-launch distance.

SO IN CONCLUSION: Assuming earth's gravitation pull remains constant
from here to the sun (which it doesn't, but work with me here), the
body of a 132 lb person will produce enough energy in nuclear fusion
of hydrogen, oxygen and carbon (assuming perfectly ideal conditions
and reactions) to propel an elephant from earth to the sun.

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