# At any given instant, how fast might an electron be moving?

1. Mar 1, 2016

### Potatochip911

1. The problem statement, all variables and given/known data
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?
2. Relevant equations
3. The attempt at a solution

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

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?

2. Mar 2, 2016

### Staff: Mentor

The point of the exercise is to show that the more a particle is constrained in space, the higher the probability that it will go fast, for which you need to use the uncertainty principle.

But the problem is not asking for the minimum speed. In a sense, it is asking for for the most probable speed, of which the standard deviation is a proxy.

3. Mar 2, 2016

### Potatochip911

Perhaps the non physics related statistics class I've taken is hindering my comprehension of this because normally the standard deviation is not a great predictor of the actual value.

4. Mar 2, 2016

### Staff: Mentor

Indeed. But here, the average momentum is zero, so the standard deviation will give you an idea of the possible range of momenta.

5. Mar 2, 2016

### Potatochip911

Ok that's interesting. Then $\sigma_p^2=\frac{\sum(p-\overline{p})^2}{n}=\frac{\sum p^2}{n}=\frac{\sum m^2v^2}{n}=\frac{m^2}{n}\sum v^2\Longrightarrow \sigma_p=m\sqrt{\frac{\sum v^2}{n}}$

So I can see why $\sigma_v$ would make sense for an approximate speed $v$. I managed to find a portion in my textbook where they mention that they're leaving out a factor of 1/2 for the particle in a box because they're interested in the magnitude and my answer above is off by a factor of 1/2 so this is making a lot more sense now.