Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How is heat generated in decay?

  1. Dec 5, 2014 #1
    I'm trying to tinker with a little home radioactive material, like Americium-241.

    I'm taking all the precautions for shielding, even though Americium isn't that dangerous behind any surface.

    So I understand the physics behind WHY radioactive decay generates heat, I'm wondering, though, what the Americium decay particles (or any radioactive material) need to be interacting with to heat?

    Is there a certain material specific for Americium that will allow the decay process to generate heat? Otherwise, why isn't the Americium in my smoke detector heating up?

    Sorry if I didn't explain it properly, hope someone can help! Thanks!
  2. jcsd
  3. Dec 5, 2014 #2

    Doug Huffman

    User Avatar
    Gold Member

    Do a mass specific activity on your Am-241 and know that the kinetic energy of the decay products is deposited in the surrounding materials. Its decay heat is 114 Watts kg^-1 of 2He4
  4. Dec 5, 2014 #3


    User Avatar
    Science Advisor

    The energy is in the form of high energy (very fast) alpha particles. These slow down by colliding with anything around, therefore transferring energy. Heat is related to how fast the particles are moving.
  5. Dec 5, 2014 #4


    User Avatar
    2017 Award

    Staff: Mentor

    Does not matter, everything will convert most of the energy to heat. Ionizations, electronic excitations, kinetic energy to transferred to nucleons or electrons, photons... all different steps towards thermal energy. Some energy will get lost to the creation of crystal defects.
    It does, but the activity is so small you don't notice it.
  6. Dec 5, 2014 #5
    So just me trying to wrap my head around this :olduhh:-no matter what the Americium is surrounded by, the alpha decay will cause the surrounding material to heat up? And the amount of heat produced is correspondent to the mass of the sample of Americium.
  7. Dec 6, 2014 #6


    User Avatar

    Staff: Mentor

    Strictly speaking, it's proportional to the number (and energy) of the alpha particles emitted. But that number is proportional to the mass, so it comes down to the same thing.
  8. Dec 6, 2014 #7
    Most of the energy of an alpha particle goes directly to heat as it is slowed down. A fraction, usually small, goes to other things - emitted as scintillation or ionized air glow from the atoms, molecules and ions excited by alpha particles (and would be turned to heat if that light is absorbed somewhere), or breaks chemical bonds that recombine to weaker ones, like nitrogen oxides and ozone in air, hydrogen, oxygen and hydrogen peroxide in water, crystal defects in solids (in which case that energy can be finally turned to heat when the products are reacted back, or defects annealed).

    The heat emitted in unit of time is independent on where the radioisotope is or what is around. What differs is how it can be dissipated. If the americium is in free and flowing air, the heat is released in a space several cm across, and then blown away, so the accumulation is hard to detect. In solids, the heat is released in the short free path of the alpha particle, but the solid may be a good conductor of heat (if a metal) or be right next to cooling air (if the isotope is on air surface). But if the same amount of radioisotope is surrounded by solid on all sides that is a poor heat conductor (ceramic, plastic...), you might notice the heat build up over time.
  9. Dec 6, 2014 #8

    Doug Huffman

    User Avatar
    Gold Member

    An ionizing smoke detector contains about 0.3 µg Am241, equivalent to 1 µCi or 37 kBq. At 100 Watts kg^-1, 0.3 µg will not provide sensible heat.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook