The time-energy uncertainty principle permits one to create particles in a virtu

Question

The time-energy uncertainty principle permits one to create particles in a virtual form for brief times. These are particles the production of which violates energy conservation, but only for times short enough that E t ¯h. During this short time, the virtual particles can propagate and interact, but then they must be annihilated again, so that, for times larger than t, energy is conserved. Suppose that a virtual particle for which m c2 = 306 MeV is created, and during the time of its existence, it travels with a speed close to the speed of light. Roughly how far will the particle travel during its brief existence? (That is, how far will this virtual particle exert its influence?) The value of ¯h is 1.06 × 1034 J · s. Answer in units of fm.

Explanation / Answer

You know the guiding formula for this problem is the Uncertainty Principle,
E * t / 2
Where E is the uncertainty in energy, t is the uncertainty in time, is equal to Planck’s constant over 2 * pi.

You are given E, this is the energy of the virtual particle.
E = 306 MeV

You know that the produce of the uncertainties in energy and time must be greater than (or equal to) h-bar over 2, which is jus a number, a very small number (with units of Joules seconds).

We can solve for what t needs to be so that the uncertainty is at its minimum,
t = / (2 * E)
t = (6.63 E-34 J s) / (4 * Pi * (306 MeV))
Converting 316 MeV into units of Joules, we get,
E = 4.98* 10-11 Joules
Plugging all of this in, we find that the minimum time uncertainty must be,
10.77 *10-25 seconds
Which is an incredibly short time.

If the particle was traveling at the speed of light (about 3 *108 m/s) for a length of time equal to the above calculated t it would travel a distance of….

d = v * t
(Since it is not accelerating)
v = c = 3 *108 m/s

d = (3 *108 m/s) * (10.77 *10-25 s)
d = 3.23 *10-16 meters =0.323 fm

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