## 1B.25

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

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Kimberly Koo 2I
Posts: 99
Joined: Sat Aug 17, 2019 12:17 am

### 1B.25

I don't get how to solve this problem. It states that, "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

Jiyoon_Hwang_2I
Posts: 101
Joined: Sat Sep 14, 2019 12:17 am

### Re: 1B.25

For the first part of the question you would use the formula Δp ≥ h/4πΔx. Plug in Planck's constant for h and 350*10^-12 m for Δx. After plugging in all of the values you should get an answer close to 1.51*10^-25 kg*m*s^-1. Then, use the formula Δv = Δp/m (this formula is derived from the equation p = mv). Plug in 1.51*10^-25 kg*m*s^-1 for Δp and the mass of an electron for m (9.1*10^-31 kg). You should get an answer of 1.65*10^5 m/s.

Asia Yamada 2C
Posts: 41
Joined: Wed Sep 30, 2020 9:36 pm

### Re: 1B.25

The diameter represents the uncertainty in position because the electron is confined to the space within the lead atom. Since you know Δx and ℏ (“h bar”), you can plug these values into Heisenberg’s Indeterminacy Equation (Δp x Δx ≥ 1/2 ℏ) to solve for Δp. Because the mass of an electron is 9.109x10^-31 kg and Δp is mass x velocity, you divide Δp by that mass to find Δv.

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