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### Speed of an Electron (hw 1.43)

Posted: Sun Oct 04, 2015 1:52 pm
What is the minimum uncertainty in the speed of an electron confined to within a lead atom of diameter 350. pm? How would you model the atom as a one-dimensional box with a length equal to the diameter of the actual atom?

### Re: Speed of an Electron (hw 1.43)

Posted: Sun Oct 04, 2015 3:57 pm
So to solve this problem, you first use Heisenberg's Indeterminacy Equation, which looks like the following:

Delta p x Delta x >= h/(4pi)

The question gives us the uncertainty in position (Delta x) by stating the diameter of the lead atom, 350 pm. Because we want to keep our units to meters, we make this 3.50 x 10^-10 m.

Now, we only have one unknown: Delta p, or the uncertainty in momentum. To find this, we plug all known values into the Heisenberg equation and solve for Delta p like this:

Delta p = 6.63x10^-34/ (4pi x 3.50x10^-10m)

Once you get p, plug that value into the equation Delta p = mass x Delta velocity to get your answer.

Note: Make sure you are careful about using parentheses, especially when doing so much multiplication and division, because not doing that messed me up a few times!

### Re: Speed of an Electron (hw 1.43)

Posted: Fri Oct 19, 2018 12:19 pm
how would you model the atom in the second part of this question?

### Re: Speed of an Electron (hw 1.43)

Posted: Fri Oct 19, 2018 1:17 pm
you could draw a square with the perimeter of the square representing the shell or orbit of an atom and putting an electron there, which would symbolize a potential place where the electron can be.

### Re: Speed of an Electron (hw 1.43)

Posted: Tue Oct 23, 2018 6:08 pm
What would be the mass for delta p = mass * delta v??

### Re: Speed of an Electron (hw 1.43)

Posted: Tue Oct 23, 2018 6:20 pm
You would just use the mass of an electron, which is m = 9.109 383 × 10−31 kg. Hope this helps!

### Re: Speed of an Electron (hw 1.43)

Posted: Thu Nov 28, 2019 2:02 pm
Hi, I have been solving for delta p with this equation: (6.62607x10^-34)/(4pi)(350x10^-12) and then using this answer in the equation delta P=mv. However, my answer continues to be 1.63x10^6. What could I be doing wrong?