## 1B #25

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

Deena Doan 2F
Posts: 100
Joined: Sat Aug 17, 2019 12:17 am

### 1B #25

Hi,
Can someone please explain this problem to me? I don't understand it.

What is the minimum uncertainty in the speed of an electron confined within a lead atom of diameter 350. pm? Model the atom as a one-dimensional box with a length equal to the diameter of the actual atom.

Thank you!

ariaterango_1A
Posts: 57
Joined: Wed Sep 18, 2019 12:15 am

### Re: 1B #25

Hi! So the uncertainty equation states that uncertainty in momentum (delta p) * the uncertainty in position (delta x) is greater than or equal to h/4pi. So first we convert 350 pico-meters to meters which is your delta x. Then you calculate (6.626 x10^-34)/ (4)*(3.14)*(350 x10^-12m) to find the uncertainty in momentum Then we use the equation uncertainty in momentum (delta p)= mass (of an electron) * uncertainty in velocity to calculate the velocity of 1.65x10^5 m.s^-1.

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