problem 43 6th edition


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

problem 43 6th edition

Postby Vanessa Reyes_1K » Thu Oct 18, 2018 9:15 pm

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

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

Re: problem 43 6th edition

Postby Mukil_Pari_2I » Thu Oct 18, 2018 9:35 pm

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
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Re: problem 43 6th edition

Postby Fionna Shue 4L » Sun Oct 21, 2018 7:50 pm

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
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Re: problem 43 6th edition

Postby Sam Kelly 1K » Tue Oct 23, 2018 12:21 pm

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

Vanadium Wang 4H
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Joined: Fri Sep 28, 2018 12:19 am

Re: problem 43 6th edition

Postby Vanadium Wang 4H » Sat Oct 27, 2018 10:33 pm

Yes, the diameter of the lead atom is given in the problem as 350 pm or 350 x 10^-12 m


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