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Homework Help: Radioactive Nuclear Polonium Battery Question

  1. Mar 18, 2010 #1

    The α-emitter 210, 84 Po is used in a nuclear battery. The kinetic energy of the emitted alpha particles can be worked out from energy and momentum conservation. It is converted into electrical energy at 5.0% efficiency.

    (a) Determine the initial power output (in W) of the battery if it contains 320 g of 210, 84 Po. Look "www.google.com/search?hl=en&q=210+Po+nuclear+decay"[/URL] (for instance) for relevant data.

    (b) Determine the power output (in W) of the battery in (a) after 1 year of operation.

    [RIGHT][8 marks][/RIGHT]


    not sure if anyone agrees with me on this one, but this work is worth 40% of the entire module and they have the cheek to link me to a google search and effectively tell me to find all the necessary values myself, i wasn't impressed...especially when you think that you might have the knowledge to do the question and just not know where to find reliable values from...

    anyway on with my answer

    (a) well i would firstly use [B]kinetic energy = 1/2 mv[SUP]2[/SUP][/B] where m=mass of an alpha particle and v=emission speed then times that by amount emitted per second (activity) to get an answer in joules per second (ie Watts) then times by 0.05 for an efficiency factor of 5%

    (b) find the half life of 210, 84 Po and use [B]N = N[SUB]0[/SUB] e[SUP]-[tex]\lambda[/tex] t[/SUP][/B] to find the number of nucleons left after one year then use [B]R=[tex]\lambda[/tex] N[/B] for the activity after one year, then simply repeat the method for question (a)...

    i appreciate that these might be the right methods but i posted this in order to give you an idea of what values i'm searching for, if im wrong in the solutions above please say.

    any feedback would be much appreciated, thanks guys
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Mar 18, 2010 #2


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    Everything you said looks correct to me. Probably when you look up the alpha emission of Po-210, you will find the alpha emission energies directly, so you won't need to use E=1/2 mv^2.
  4. Mar 18, 2010 #3
    Thanks for the reply phyzguy i was thinking i needed loads of extra info but now i realise the only bit of information about Po 210 that I could find online would be its half life as the activity and decay constant etc etc etc depend on how much of the stuff is left in this case 320g
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