## 1B.25

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

Midori Hupfeld 3D
Posts: 55
Joined: Fri Aug 30, 2019 12:17 am

### 1B.25

Can someone explain how to approach 1B #25? I looked at the solution and it rearranged an inequality equation that does not look like the one we discussed in lecture. It says that the uncertainty in momentum times the uncertainty in velocity is equal to 1/2 times h with a line through the top.

Diana_Diep2I
Posts: 130
Joined: Sat Aug 17, 2019 12:17 am

### Re: 1B.25

They just used the shorthand with the slashed h. Otherwise, the problem setup is the same. 1)You set up Heisenberg's principle 2)divide change in position aka the diameter of the atom 3)it asks for uncertainty in velocity, so you change the p in momentum to mΔv 4)divide the mass by both sides to get Δv on the left only.

at the end, you should get Δv≥h/4πΔxm. This key just broke up the steps and used the shorthand.

William Chan 1D
Posts: 102
Joined: Sat Sep 14, 2019 12:15 am

### Re: 1B.25

h bar, or ℏ, is h/2π.