A proton is confined in an atomic nucleus of size 2 fm. Determine the fundamental limit on
the uncertainty of any prediction for the measurement of the momentum of the proton in the
nucleus.
The uncertainty in proton location is
deltax = +/- 1 fm = 10^-15 m
The uncertainty in x-momentum is
deltaPx = (h/4pi)/deltax = 5*10^-20 kg m/s
where h is Planck's constant.
Your book may eliminate the 1/(4 pi) factor. It depends upon how the uncertainty is defined. The h/4 pi value is the product of standard deviations.
Remember that there are uncertainties in three directions of momentum. The total momentum uncertainty is a factor of sqrt3 higher.
I would not be surprised if your book's answer is up to a factor of ten different.
No wonder they call it the uncertainty principle. :-)
I don't understand why you've used 4 pi in your answer, i assume your rearranged the uncertainty principle equation for delta Px, but isnt the reduced plancks constant h/ 2pi instead of 4pi ?
The fundamental limit on the uncertainty of any prediction for the measurement of the momentum of a particle can be determined using the Heisenberg uncertainty principle. According to this principle, there is an inherent uncertainty in simultaneously measuring the position and momentum of a particle.
The uncertainty principle is described by the equation: Δx * Δp ≥ h/4π, where Δx represents the uncertainty in position, Δp represents the uncertainty in momentum, and h is the Planck constant.
In this case, we are interested in finding the uncertainty in momentum (Δp) of a proton confined in an atomic nucleus of size 2 fm (femtometers), which is equivalent to 2 * 10^(-15) meters.
To determine the uncertainty in position (Δx), we can take the size of the nucleus, which is 2 fm, as a measure of the uncertainty in position. Therefore, Δx = 2 fm = 2 * 10^(-15) meters.
Now, we can rearrange the uncertainty principle equation to solve for Δp:
Δp ≥ h/4πΔx
Plugging in the values, we get:
Δp ≥ (6.626 x 10^(-34) J s) / (4π * 2 * 10^(-15) m)
Simplifying further:
Δp ≥ (6.626 x 10^(-34) J s) / (8π x 10^(-15) m)
Now we can calculate the value of Δp:
Δp ≥ (6.626 x 10^(-34))/(8π x 10^(-15)) J s/m
Using a calculator:
Δp ≥ 2.09 x 10^(-19) kg m/s
Therefore, the fundamental limit on the uncertainty for the measurement of the momentum of the proton in the nucleus is approximately 2.09 x 10^(-19) kg m/s.