- #1

Potatochip911

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## Homework Statement

This is problem 5-42 from Modern Physics by Tipler & Llewellyn. Neutrons and protons in atomic nuclei are confined within a region whose diameter is is about ##10^{-15}##m. At any given instant, how fast might an individual proton or neutron be moving?

## Homework Equations

3. The Attempt at a Solution [/B]

##m\approxeq 1.67\times 10^{-27}kg##

Looking at this question I would assume it is a particle in a one dimensional box with length ##10^{-15}##m. Therefore we can find the minimum speed for a proton or neutron by using the equations:

(1) ##E_{kmin}=\frac{\hbar^2}{2mL^2}##

(2) ##\frac{1}{2}mv^2=E_{kmin}##

which results in ##v_{min}=\sqrt{\frac{\hbar^2}{m^2L^2}}=\frac{\hbar}{mL}##

The solution manual however uses the uncertainty principle: ##\Delta x\Delta p\approx \frac{\hbar}{2}## then assuming the particle to be non-relativistic we have ##\Delta p=m\Delta v## hence we obtain

##\Delta v=\frac{\hbar}{2m\Delta x}##

This seems incorrect to me, the question is asking how fast a proton or neutron

*might*be moving not the uncertainty in it's speed especially considering the uncertainty principle can be stated as

##\sigma_x\sigma_p\geq \frac{\hbar}{2}## where ##\sigma_x## is the standard deviation in position and ##\sigma_p## is the standard deviation in momentum. Why would we be using symbols that are defined as standard deviations to determine the minimum speed?