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

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

    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 [Broken]
    Last edited by a moderator: May 2, 2017
  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]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook