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### HW #1.43

Posted: **Thu Apr 26, 2018 11:56 am**

by **Fiona Grant 1I**

What is the minimum uncertainty in the speed of an electron confined to within a lead atom of diameter 350. pm?

When I calculated the answer to this problem, I got an answer of 1.51 x 10^-25 m/s. However, apparently the solution should be 1.65 x 10^5 m/s. How am I getting a value that is so off?

### Re: HW #1.43

Posted: **Thu Apr 26, 2018 11:57 am**

by **Natalie Noble 1G**

You got indeterminacy of momentum which is kg.m/s and speed is m/s so you need velocity

Indeterminacy of momentum doesn’t tell you what you were asked for, you are asked to find the speed so that is velocity, so you need to find the indeterminacy of velocity

Indeterminacy of momentum = 1.507 x 10^-25 kg.m/s

p=m*v

v=m/p

v= 9.109383kg/1.507 x 10^-25 kg.m/s

v= 1.65 x 10^5 m/s

### Re: HW #1.43

Posted: **Tue May 01, 2018 10:39 pm**

by **Andrew Berard 1F**

Just a correction,

v=p/m

So it would be 1.507 x 10^-25 km/s / 9.109383 kgm/s

which gets you 1.65 x 10^5 m/s

### Re: HW #1.43

Posted: **Sun May 06, 2018 1:09 pm**

by **304984981**

just remember P=mv where m is a constant which is the mass of a electron, 9.11E-31 kg

And according to Heisenberg Indeterminacy (Uncertainty) Equation: px=h/4pi, and because p=mv, so mvx=h/4pi

put in all the variable, where m=9.11E-31, x=350pm=350E-12 m, h=6.63E-34,

we get the final answer 165469.2289 which u can write as 1.65E5

### Re: HW #1.43

Posted: **Tue May 08, 2018 8:58 pm**

by **Natalie_Martinez_1I**

How did you find the uncertainty in the momentum for this problem?

### Re: HW #1.43

Posted: **Thu Jun 14, 2018 12:59 pm**

by **105012653 1F**

^^ I had the same question.. still lost on that

### Re: HW #1.43

Posted: **Sat Oct 13, 2018 8:13 pm**

by **zoepamonag4C**

On the solutions manual, one of the values is h_bar= 1.054 457 * 10^-34 J/s. Does anyone know where they are getting this number from?

### Re: HW #1.43

Posted: **Sun Oct 14, 2018 2:34 pm**

by **404905747**

I had the same question and read through the replies but am still a little confused. Can someone maybe rephrase how they got to the answer?

### Re: HW #1.43

Posted: **Sun Oct 14, 2018 9:55 pm**

by **ryanhon2H**

h bar is said in the textbook to be h/2π, where h is Planck's constant

The uncertainty principle equation is

(∆p)(∆x) = .5(h bar) or (∆p)(∆x) = h/4π

∆p is the momentum uncertainty and ∆x is the position uncertainty. ∆x is given in the problem, which is 350. pm or

3.50 x 10^{-10} meters. Plug it into the equation to solve for ∆p

Then, to find the minimum uncertainty for velocity you have to use the relationship

p = mv

to get

∆p = m(∆v)

where m is the mass of the electron. Since you just solved for ∆p and know m, you can solve for ∆v and get the answer.