# Homework Help: Application of Heisenberg Uncertainty Principle

1. Apr 20, 2015

### Raghav Gupta

1. The problem statement, all variables and given/known data
A parallel beam of electrons traveling in x- direction falls on a slit of width d.
If after passing the slit, an electron acquires momentum py in the y direction then for a majority of electrons passing through the slit
A. |py|d ≈ h
B. A. |py|d > h
C. |py|d < h
D. |py|d >> h

2. Relevant equations
Δx Δp = h/4π

3. The attempt at a solution
I only know Hiesenberg uncertainty equation
How to deal here with inequalities?

2. Apr 20, 2015

### gleem

First the Heisenberg uncertainty relation should be written ΔxΔp ≥ h/4π . What does this expression imply?

3. Apr 20, 2015

### Raghav Gupta

This implies that we have uncertainty in position and momentum calculation of an electron.
How to apply that in this question ?

4. Apr 20, 2015

### gleem

When the electron passes through the slit of width d what are you doing? Using the Heisenberg Uncertainly relation what statement can you make about the lateral momentum

5. Apr 20, 2015

### Raghav Gupta

Δp >= h/4πd
Is that correct?

6. Apr 20, 2015

### gleem

Yes, and what does this suggest?

7. Apr 20, 2015

### Raghav Gupta

Sorry, I think it should be Δp>= h/4πΔd ?
There should be uncertainty involved.

8. Apr 20, 2015

### gleem

No. you are localizing the electron in a space of width d which is the significance of Δx.

9. Apr 20, 2015

### Raghav Gupta

That uncertainty in momentum is greater?

10. Apr 20, 2015

### gleem

Yes, or the uncertainty is not less than h/4πd.

I have to do something so I will be away for a few hours sorry.

11. Apr 20, 2015

### Raghav Gupta

How does that relate to options?

Think you will reply after some hours.

12. Apr 20, 2015

### gleem

You should be able to exclude two immediately>

13. Apr 20, 2015

### Raghav Gupta

Do that mean certainty is less then h/4πd ?
Which is py<h/4πd ?
Why absolute value is being considered of momentum in options and why 4π term is neglected?
This is not true , if pyd< h/4π
Then |py|d< h ?

Last edited: Apr 21, 2015
14. Apr 21, 2015

### gleem

The expression of the uncertainty principle was original suggested by Heisenberg to be ΔpΔx ≥ h. It was later formally derived and found to be ΔpΔx ≥ h/4π.

It seems to me that the two forms are mixed here. First even is they are mixed A and C are not consistent with either one. But if Δp ≥ h/4πd you cannot say Δp > h/d let alone Δp >> h/d. So to make any sense out of the choices you have to use Heisenberg's original formulation ΔpΔx ≥ h.

The absolute value is used because momentum and position are vector quantities and only the magnitudes of these quantities are of interest.

15. Apr 21, 2015

### Raghav Gupta

So option C is correct, by original formulation also and also by modern formulation?
If pyd < h/4π
Then 4πpyd < h
Which definitely implies pyd <h
So no matter what formulation we take ,we only arrive at this answer ?

16. Apr 21, 2015

### gleem

If Δpd ≥ h/4π then 4π Δpd ≥ h. But this does not imply universally that Δpd < h.

17. Apr 21, 2015

### Raghav Gupta

Is this correct?

18. Apr 21, 2015

### gleem

No. you reversed the inequality sign.

19. Apr 21, 2015

### Raghav Gupta

But I am taking certainty.
When we remove delta sign , inequality changes?

20. Apr 21, 2015

### gleem

It is my understanding that the uncertainty relationship makes no statement about the actual momentum of a particle.