1B.25

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FDeCastro_1B
Posts: 54
Joined: Thu Jul 25, 2019 12:16 am

1B.25

Postby FDeCastro_1B » Mon Oct 14, 2019 4:27 pm

What is the minimum uncertainty in the speed of an electron confined to within a lead atom of diameter 350. pm? Model the atom as a one-dimensional box with a length equal to the diameter of the actual atom.

I used the Heisenbeg Indeterminacy Equation and still did not get the same answer as what's shown in the solutions manual. What am I doing wrong?

Abby Soriano 1J
Posts: 103
Joined: Sat Aug 24, 2019 12:16 am

Re: 1B.25

Postby Abby Soriano 1J » Mon Oct 14, 2019 4:36 pm

If you look on the class website, there's actually an error in the solutions manual for that problem. They made delta v = 5 m/s rather than 10 m/s as it should be (the velocity given was 5 m/s +- 5 m/s, so that would range from 0-10 m/s). The correct answer for delta x = 6.7 x 10-37 m (hopefully you got that!)

FDeCastro_1B
Posts: 54
Joined: Thu Jul 25, 2019 12:16 am

Re: 1B.25

Postby FDeCastro_1B » Mon Oct 14, 2019 4:40 pm

Abby Soriano 1H wrote:If you look on the class website, there's actually an error in the solutions manual for that problem. They made delta v = 5 m/s rather than 10 m/s as it should be (the velocity given was 5 m/s +- 5 m/s, so that would range from 0-10 m/s). The correct answer for delta x = 6.7 x 10-37 m (hopefully you got that!)


thank you!!

Abby Soriano 1J
Posts: 103
Joined: Sat Aug 24, 2019 12:16 am

Re: 1B.25

Postby Abby Soriano 1J » Mon Oct 14, 2019 4:57 pm

FDeCastro_1B wrote:
Abby Soriano 1H wrote:If you look on the class website, there's actually an error in the solutions manual for that problem. They made delta v = 5 m/s rather than 10 m/s as it should be (the velocity given was 5 m/s +- 5 m/s, so that would range from 0-10 m/s). The correct answer for delta x = 6.7 x 10-37 m (hopefully you got that!)


thank you!!


Oh wait I was thinking of a different problem and accidentally gave you wrong information sorry!

In order to find the uncertainty in speed, you have to first figure out the uncertainty in momentum with the given uncertainty in position. If you plug in 350. pm (converted into m) into Heisenberg's eqution for delta x, you can then figure out delta p. Upon finding delta p, just divide it by the mass of an electron (9.11 x 10-31 kg) in order to find your delta v (since momentum is just mass x velocity).

Hope this helps and sorry for the confusion!


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