## problem 43 6th edition

Vanessa Reyes_1K
Posts: 60
Joined: Fri Sep 28, 2018 12:28 am

### problem 43 6th edition

For problem 43 in the 6th edition book, what is meant by "minimum uncertainty"? How do you solve for it?

Mukil_Pari_2I
Posts: 87
Joined: Fri Sep 28, 2018 12:29 am

### Re: problem 43 6th edition

The minimum uncertainty is change in velocity or delta (the triangle) v. In this case, delta v would be 1.65*10^5 m/s.

Fionna Shue 4L
Posts: 59
Joined: Fri Sep 28, 2018 12:18 am

### Re: problem 43 6th edition

To solve for this, we would use the equation ∆x∆p = h/4π. For an electron, ∆p = m∆v, and therefore, m∆x∆v = h/4π. The uncertainty on position of an electron would be the diameter of the atom so ∆x = 350 pm. "m" is the mass of the electron which is 9.109*10^-31 kg and h=6.626*10^-34 J.s

Sam Kelly 1K
Posts: 30
Joined: Fri Sep 28, 2018 12:16 am

### Re: problem 43 6th edition

Fionna Shue 1L wrote:To solve for this, we would use the equation ∆x∆p = h/4π. For an electron, ∆p = m∆v, and therefore, m∆x∆v = h/4π. The uncertainty on position of an electron would be the diameter of the atom so ∆x = 350 pm. "m" is the mass of the electron which is 9.109*10^-31 kg and h=6.626*10^-34 J.s

How do you determine the diameter of the atom in this case? Is it given in the problem