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### Topic: how many "radioactive" people would be required to power a 100 W light bulb?  (Read 13468 times)

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#### nakita_66

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##### how many "radioactive" people would be required to power a 100 W light bulb?
« on: March 23, 2006, 10:14:25 AM »
this question was given in lecture, and i have no idea how to go about it any help would be amazing.
thanks,

If all of the energy from the decay of 14C in your body could be captured and converted into electricity, how many "radioactive" people would be required to power a 100 W light bulb?

« Last Edit: March 23, 2006, 12:43:40 PM by Mitch »

#### Grejak

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##### Re:how many "radioactive" people would be required to power a 100 W light bulb
« Reply #1 on: March 23, 2006, 07:08:39 PM »
About 1.5x1012.  Assuming that a human is 75 kg, 18% carbon, the 12C/14C ratio is the same as it is in the atmosphere, that antineutrinos do not exist (or that you can use them to power light bulbs) and that I can properly do math.

Either way, it is a lot.
« Last Edit: March 23, 2006, 07:09:15 PM by Grejak »

#### nakita_66

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##### Re:how many "radioactive" people would be required to power a 100 W light bulb
« Reply #2 on: March 24, 2006, 12:16:52 AM »
if its not too much trouble could you show me the calulation process? thanks

#### Grejak

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##### Re:how many "radioactive" people would be required to power a 100 W light bulb
« Reply #3 on: March 24, 2006, 11:00:57 PM »
Well, the question is asking how many people it will take to give off 100 Watts of power.  Since each watt is a joule/second, what you need to do is to find out how many joules of energy each person gives off per second from 14C.  In order to figure this out you need a few pieces of information:

1. What energy is released in each 14C decay.
2. How many decays per second occur for each person.

For part 1, the energy released in beta decay is the Q value of the reaction which can be determined from the mass defects: 3.0199 MeV - 2.8634 MeV = 0.1565 MeV per 14C decay.  You can then convert that number into Joules.

For the second question, you need to make some assumptions.  The ones I made were that each person weighs 75 kg, of which 18% of that weight was due to carbon.  I also assumed that the 12C:14C ratio in the human body is the same as it is in the atmosphere or around 1012:1.  Based off of that information you can determine the number of 14C atoms in a human body.  Once you know how many 14C atoms the body contains, you can determine the activity thanks to the relationship: Activity = Number of 14C atoms * ln(2)/Half-life.  You should get something around 2.597E+3 decays per second.

This allow you to determine how much energy a human gives off due to 14C decay each second.  Take the 100 Watts and divide it by the Joules/second from a human and you will be able to figure out how many humans you need in order to power a 100 Watt light-bulb.

#### nakita_66

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##### Re:how many "radioactive" people would be required to power a 100 W light bulb
« Reply #4 on: March 26, 2006, 09:55:12 AM »
thanks soo much.

#### nakita_66

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##### Re:how many "radioactive" people would be required to power a 100 W light bulb
« Reply #5 on: March 28, 2006, 07:57:56 PM »
i still cant seem to get teh right calulations im not sure if i can calculate the number of atoms in a person, also when you use the activity equation do you use half life in years or seconds?

#### jdurg

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##### Re:how many "radioactive" people would be required to power a 100 W light bulb
« Reply #6 on: March 31, 2006, 11:25:10 AM »
Because a Watt is  1 Joule per Second, to get the calculations correct you'll need to use the half-life in terms of seconds, not years.
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