## HW #1.43

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

Fiona Grant 1I
Posts: 30
Joined: Fri Apr 06, 2018 11:03 am

### HW #1.43

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?

Natalie Noble 1G
Posts: 30
Joined: Thu Feb 01, 2018 3:02 am
Been upvoted: 1 time

### Re: HW #1.43

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

Andrew Berard 1F
Posts: 21
Joined: Fri Apr 06, 2018 11:01 am

### Re: HW #1.43

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

304984981
Posts: 28
Joined: Fri Apr 06, 2018 11:05 am

### Re: HW #1.43

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

Natalie_Martinez_1I
Posts: 30
Joined: Mon Feb 26, 2018 3:00 am

### Re: HW #1.43

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

105012653 1F
Posts: 30
Joined: Fri Apr 06, 2018 11:02 am

### Re: HW #1.43

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

zoepamonag4C
Posts: 30
Joined: Fri Sep 28, 2018 12:28 am

### Re: HW #1.43

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?

404905747
Posts: 29
Joined: Fri Sep 28, 2018 12:17 am

### Re: HW #1.43

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?

ryanhon2H
Posts: 60
Joined: Fri Sep 28, 2018 12:28 am

### Re: HW #1.43

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.