Discuss the prohibition of E = 0 for a particle trapped in a box L wide in terms of the uncertainty principle. How does the minimum momentum of such a particle compare with the momentum uncertainty required by the uncertainty principle if we take Delta*x =L?

The prohibition of E = 0 for a particle trapped in a box L wide is a consequence of the Heisenberg uncertainty principle, which states that the product of the uncertainties in position and momentum of a particle must be greater than or equal to a certain constant (h/4π).

If we consider a particle trapped in a box of width L, the uncertainty in position, Δx, can be taken as L. This is because the particle is known to be somewhere within the box, but its exact position is uncertain.

The uncertainty principle then gives us Δp ≥ h/(4πL), where Δp is the uncertainty in momentum. This means that the particle must have a minimum momentum uncertainty of h/(4πL).

Now, the energy of a particle is given by E = p²/2m, where p is the momentum and m is the mass of the particle. If E = 0, this would imply that p = 0. However, this would contradict the uncertainty principle, which requires that Δp ≥ h/(4πL). Therefore, E cannot be 0 for a particle trapped in a box L wide.

In terms of how the minimum momentum of such a particle compares with the momentum uncertainty required by the uncertainty principle, the minimum momentum is actually equal to the momentum uncertainty. This is because the uncertainty principle sets a lower limit on the product of the uncertainties in position and momentum, and in this case, we have taken the uncertainty in position to be as large as possible (L), so the uncertainty in momentum must be as small as possible, i.e., the minimum momentum.