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Smoke Detectors

  1. Jan 25, 2010 #1
    Most smoke detectors contain an isotope of Americium (241Am), the nuclei of which decay by emitting an alpha particle. When a nucleus of Americium emits an alpha particle the principle of conservation of momentum and the principle of conservation of energy are obeyed.

    Show that the ratio (KE of alpha particle)/(KE of resulting nucles) is about 60 and briefly discuss the origin of this energy.

    The answer given is:

    E = mc^2 (energy must have come from mass)

    Total mass after is a little less than before

    I don't get what this means??
  2. jcsd
  3. Jan 25, 2010 #2


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    Write the equation for the decay.

    Find the energy before and the energy after using E=mc2
  4. Jan 25, 2010 #3

    Andrew Mason

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    An alpha particle consists of 2 protons and 2 neutrons so it has an atomic mass of very close to 4 amu. Americium has an atomic mass very close 241 amu. The recoiling nucleus would have an atomic mass very close to 237 amu. The ratio of the atomic mass of the recoiling nucleus would be 237/4 = 59.25.

    From conservation of momentum:

    [tex]m_{\alpha}v_{\alpha} + m_nv_n = 0[/tex]

    Work out the ratio of the energy of the alpha particle to the energy of the recoiling nucleus. You should get:

    [tex]\frac{KE_{\alpha}}{KE_n} = \frac{m_n}{m_{\alpha}} = 59.25[/tex]

    However, if you were to measure the masses of the alpha particle and recoiling nucleus, the total mass would be a small amount less than the mass of the original Americium nucleus. This 'mass defect' is the source of the energy - this mass has been converted into energy according to Einstein's equation: E = mc^2.

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