## 1B.27 Hw Help

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

Bita Ghanei 1F
Posts: 60
Joined: Thu Feb 28, 2019 12:15 am

### 1B.27 Hw Help

A bowling ball of mass 8.00 kg is rolled down a bowling alley lane at . What is the minimum uncertainty in its position?

How do you solve for the change in velocity here? It seems as though we are not given enough information.

Paul Hage 2G
Posts: 105
Joined: Thu Jul 25, 2019 12:17 am

### Re: 1B.27 Hw Help

You would use the Heisenberg Uncertainty Equation: ΔpΔx≥h/4pi. This equation is equivalent to mΔvΔx≥h/4pi. So you would now isolate Δv and sub in the values we know. We know, from the problem, that the uncertainty in position, Δx, is 5.0m. The mass is 8.00kg. h is Planck's constant, 6.626 x 10^-34 J s. Using this information, we can isolate Δv and solve.

Sion Hwang 4D
Posts: 51
Joined: Wed Sep 18, 2019 12:21 am

### Re: 1B.27 Hw Help

You use the Heisenberg's Uncertainty Equation.

It states:
ΔmΔvΔx ≥ h/4pi.
You know that the mass is 8.00 kg, h= 6.626 x 10^-34 kg*m^2*s^-1, and Δv is 10m/s. You must isolate Δx.

Hence,

Δx ≥ (6.626 x 10^-34 kg*m^2*s^-1 / 4pi) / (8kg * 10m/s), and you are left with:
Δx ≥ 6.6 x 10^-37 m.

RRahimtoola1I
Posts: 102
Joined: Fri Aug 09, 2019 12:15 am

### Re: 1B.27 Hw Help

For ∆x, I got 6.6 x 10-36m also, but the answer in the back of the book and the solution manual is 1.3 x 10-36m. Which is the correct answer?

AlyssaYeh_1B
Posts: 100
Joined: Sat Aug 17, 2019 12:16 am

### Re: 1B.27 Hw Help

RRahimtoola1G wrote:For ∆x, I got 6.6 x 10-36m also, but the answer in the back of the book and the solution manual is 1.3 x 10-36m. Which is the correct answer?

The correct answer would be 6.6 x 10-37m. There was an error in the solutions manual (Δv = 5m/s when Δv should've been 10m/s)

805307623
Posts: 99
Joined: Fri Aug 09, 2019 12:17 am

### Re: 1B.27 Hw Help

If you plug in all the variables the equation: delta x= (1/2)(h/(m*delta v)) should provide the answer straight away.