Kinetic Energy based on uncertainty


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Alexandra Albers 1D
Posts: 61
Joined: Fri Sep 28, 2018 12:18 am

Kinetic Energy based on uncertainty

Postby Alexandra Albers 1D » Sat Oct 20, 2018 11:03 am

In the post-assessment for the video module on the Heisenberg Uncertainty Equation, there is a question where you are asked to calculate the electron's uncertainty in kinetic energy and then in kinetic energy per mole of electrons. I'm not sure exactly how to do this problem, so what equations should I use to convert the uncertainty in velocity to kinetic energy?

Elaine Pham 2E
Posts: 58
Joined: Fri Sep 28, 2018 12:28 am
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Re: Kinetic Energy based on uncertainty

Postby Elaine Pham 2E » Sat Oct 20, 2018 11:35 am

The formula for kinetic energy is KE=(mv^2)/2. With this post-assessment problem, you can plug in the uncertainty of the velocity and the mass of an electron, which is 9.1 x 10^-31 kg (I just searched it up on Google), into the KE equation and get the uncertainty in kinetic energy. Keep in mind that the answer you get after plugging in the uncertainty in velocity and mass of the electron is the uncertainty in KE for just 1 electron. So to get the uncertainty in KE for a whole mole of electrons, you multiply the answer you got by Avogadro's constant.


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