## Kinetic Energy Uncertainty Conversion

$\Delta p \Delta x\geq \frac{h}{4\pi }$

Trent Yamamoto 2J
Posts: 111
Joined: Sat Aug 17, 2019 12:18 am

### Kinetic Energy Uncertainty Conversion

In the Heisenberg Indeterminacy Module, question 22 asked for the uncertainty in kinetic energy (which I was able to calculate) and the conversion of that into moles. How would I do that?

Kevin Antony 2B
Posts: 99
Joined: Sat Sep 07, 2019 12:16 am

### Re: Kinetic Energy Uncertainty Conversion

From what I gather, you know have energy per electron. We know there are 6.022 x 10^23 electrons in a mol. If you divide by Avagadro's number you should get energy per mol.

Trent Yamamoto 2J
Posts: 111
Joined: Sat Aug 17, 2019 12:18 am

### Re: Kinetic Energy Uncertainty Conversion

Kevin Antony 1J wrote:From what I gather, you know have energy per electron. We know there are 6.022 x 10^23 electrons in a mol. If you divide by Avagadro's number you should get energy per mol.

That makes sense, thank you!

britthanul234
Posts: 51
Joined: Wed Sep 18, 2019 12:21 am

### Re: Kinetic Energy Uncertainty Conversion

There are 6.022 x 10^23 electrons in a mol. If you divide by Avagadro's number you get energy per mol.