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Calculate the lifetime of the theta meson

  1. Jan 15, 2006 #1
    A newly discovered Θ meson has a rest mass energy of 1020 MeV, and electric charge of 0, and a measured energy width of 4 MeV

    a Using the uncertainty principle, calculate the lifetime of the Θ meson

    b A Θ meson at rest decays into K+ and K- mesons. Find the total KE of the kaons

    Thanks
    and if you can also help explain the uncertainty principle
     
  2. jcsd
  3. Jan 15, 2006 #2
    the uncertainty principle stems fromthe fact every particle is described by a wave packet of probability density.
    the wave-number ([tex]k=\frac{2 \pi}{\lambda}[/tex]) of this packet is the momentum.
    when you try to make a wave packet defined in space, you need to add lots of monochromatic waves with different wave-numbers togethere - and when you got only one monochromatic wave (meaning there is only one wave-number) the wave looks like a sine which comes from minus infinity and continues to infinity. meaning you got undefined coordinates for the particle..
    so the particle can be described as
    [tex]\Psi (x)=\Sigma C_ne^{ik_nx}[/tex]
    by using fourier transform from x to k (when given the shape of particle packet in space) you could see that the minimum value of [tex]\Delta x\Delta p=\frac{1}{2}[/tex]
    (when [tex]\hbar=1[/tex], not in cgs)
    energy and time are related to each other the same way as space and momentum, since a fourier transform of x to t when [tex]\Psi (x)=\Sigma C_ne^{ik_nx}[/tex] will give you the [tex]\Psi (t)=\Sigma U_ne^{iE_nt}[/tex]
    so if you know the energy width of the particle, youd know the time width.
    by [tex]\Delta E\Delta t >= \frac{\hbar}{2}[/tex] in cgs.
    if you want more info look here http://zebu.uoregon.edu/~js/21st_century_science/lectures/lec14.html
     
    Last edited: Jan 15, 2006
  4. Jan 15, 2006 #3
    wouldnt it be ^E^t>h/2pi?
     
  5. Jan 19, 2006 #4
    well, [tex]\hbar=\frac{h}{2\pi}[/tex]
    so it would be greater then [tex]\frac{h}{4\pi}[/tex]
    but [tex]\Delta x\Delta p[/tex] roughly equals to [tex]\frac{h}{2\pi}[/tex]
     
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