The Z0 boson, discovered in 1985, is the mediator of the weak nuclear force, and it typically decays very quickly. Its average rest energy i

Question

The Z0 boson, discovered in 1985, is the mediator of the weak nuclear force, and it typically decays very quickly. Its average rest energy is 91.19 GeV, but its short lifetime shows up as an intrinsic width of 2.5 GeV. what is the lifetime of this particle?

in progress 0
Philomena 4 years 2021-08-31T16:50:03+00:00 1 Answers 31 views 0

Answers ( )

  1. Euphemia
    0
    2021-08-31T16:51:54+00:00
    Reply

    Answer:

    The lifetime of the particle is  \Delta t = 2.6*10^{-25} \ s

    Explanation:

    From the question we are told that

        The average rest energy is E = 91.19 \ GeV = 91.19GeV * \frac{1.60 *10^{-10} J }{1GeV} = 1.46 *10^{-8}J

        The intrinsic width is  \Delta E =2.5eV = 2.5GeV * \frac{1.60 *10^{-10}J }{1GeV} = 4*10^{-10} J

    The lifetime is mathematically represented as

         \Delta t = \frac{h}{\Delta E}

    Where h is the Planck’s constant with a value of  1.055*10^{-34} \ J\cdot s  

    substituting values

        \Delta t = \frac{1.055*10^{-34}}{4 *10^{-10}}

         \Delta t = 2.6*10^{-25} \ s

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )