## Sapling #19

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

Dominic Benna 2E
Posts: 93
Joined: Wed Sep 30, 2020 10:09 pm

### Sapling #19

For the second part of #19 on the sapling homework, I keep getting 5.67x10^-2 m/s, but it says that it is incorrect.

This is the question for reference:
What is the minimum uncertainty in a helium atom's velocity if the position is known to be 1.4 angstroms

Julianna_flores3E
Posts: 86
Joined: Wed Sep 30, 2020 9:52 pm

### Re: Sapling #19

You would follow basically the same procedure as the top part of the problem, except you would need to use Avogadro's # to convert to atoms and then convert to KG because it is the SI unit needed for the calculations. Then you would be able to follow the uncertainty principle with the mass of the helium atom.

Andrew Yoon 3L
Posts: 87
Joined: Wed Sep 30, 2020 9:36 pm

### Re: Sapling #19

Hi, so for this question, you still use the equation (delta momentum)(delta position) > or equal to (h/4pi). However, for delta momentum, change the mass to helium. In order for helium to match with the units in the equation, first change the molar mass from g/mol to kg/mol. Next, multiply it my Avogadro's number to get atoms. Finally, you are able to solve for the minimum uncertainty in a helium atom's velocity.

Manseej Khatri 2B
Posts: 76
Joined: Wed Sep 30, 2020 9:42 pm

### Re: Sapling #19

Hi, for this question you would use (delta x)(delta p) = h/4pi where p is the uncertainty in momentum and x is the uncertainty in position. The position's uncertainty is 1.4 x 10 ^ -10 m. Delta p =m * (delta v) which is the uncertainty in velocity. So the resulting inequality is (1.4 x 10 ^-10)( mass of a helium atom * delta v) >= h/4pi. Solving for delta v, you get the minimum uncertainty.

Susanna Givan 2B
Posts: 61
Joined: Thu Feb 27, 2020 12:16 am

### Re: Sapling #19

Thank you so much; I understand this principle so much better now!