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How can Uncertainty principle be used to get speed of alpha particle

  1. Dec 8, 2009 #1
    assuming that Delta p = aprroxiamtely p, how can the uncertainty principle, be manipulated in order to achieve the speed of an alpha particle trapped in a nucleus of just say uranium 238 with proton number 92 ??? or any other nucleus....what is the main manipulation steps of the formula???would appreciate any help, thanks!

    for the uranium above ive estimated the size of the nucleus to have radius of around 7.44 fm or (7.44 x 10^-15) m if thats supposed to help, not sure
    Last edited: Dec 8, 2009
  2. jcsd
  3. Dec 8, 2009 #2
    I'm not sure how to solve this problem - just throwing some ideas out here.

    Delta p is a little more than 'approximately p'. More formally:

    Delta p is the root of { (the expectation value of (p squared) - (expectation value of p) squared }

    i.e. {<p^2> - <p>^2} if you can read latex.

    and similarly for Delta x.

    So the uncertainty relation can be understood as relating these quantities in an inequality.

    If you know the Schrodinger equation for the system you have in mind mention, you should be able to calculate some of these expectation values.

    Presumably, when the questioner asks for the speed of an alpha particle, they're asking for the expectation value of the speed.

    Speed is connected to momentum here.

    This should at least give you something to play with - until somebody cleverer comes to help.

    However there's one thing I'm very uncertain (haha) about - the uncertainty relation is merely an inequality. I've seen the uncertainty relation used to estimate properties of systems in their ground state - because we're searching for a minimal quantity - but this aspect doesn't seem to appear in the question. So I fear I'm missing something.
    Last edited: Dec 8, 2009
  4. Dec 8, 2009 #3
    First you need to estimate the ground state level of the nucleus. That is the kinetic energy, or p^2/2m. The ground state is stationary, so <p>=0. So using [tex]\Delta p=<(p-<p>)^2>^{1/2}[/tex], [tex]\Delta p=<p^2>^{1/2}=(2m<H>)^{1/2}[/tex].

    So from the uncertainty relation [tex]\Delta p=(2m<H>)^{1/2}>\frac{\hbar}{2 \Delta x} [/tex]

    or [tex]<H> > \frac{\hbar^2}{8m (\Delta x)^2}[/tex]

    So what you know is that the potential has a value at least equal to this ground state energy, in order to be confined. So to escape the potential, the particle must have at least this energy. When it is free of the nucleus, it'll still have this energy relative to the nucleus. So at the very least, the kinetic energy is equal to [tex]\frac{\hbar^2}{8m (\Delta x)^2}[/tex], but it is actually more relative to infinity where it is free instead of relative to the nucleus. So set p^2/2m equal to [tex]\frac{\hbar^2}{8m (\Delta x)^2}[/tex], and solve for p:

    [tex]p=\frac{\hbar}{2 (\Delta x)} [/tex]

    if the size of the nucleus is 10^-14, then the momentum is 3.3*10^(-20). If the alpha particle has mass 6.6*10^(-27), then the velocity is 1/2*10^7 m/s. I think this is close enough to the speed of light (if you had used 10^-15 meters for the nucleus instead of 10^-14 then you'd get there) where you might have to use relativity to get the speed, which means you'd set (c^2p^2+m^2c^4)^(1/2) equal to the energy, and solve for p, and then use the relativistic relationship between p and v to solve for v.

    Anyways, if I didn't make a mistake, this is the long explanation of how to use the uncertainty principle.
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