# Measuring the wavelength and position of a photon

1. Oct 14, 2007

### zing478

1. The problem statement, all variables and given/known data
A wavelength measurement gives Lambda= 6000 x 10^-10m with an accuracy of one part in a million. What is the minimum uncertainty in the position of the photon?

2. Relevant equations
deltaP*deltaX > h/4pi

p = h/lambda

3. The attempt at a solution
In order to solve, I know deltaP has to be expressed in terms of deltaLambda. Delta lambda is (6000 x 10^-10)(1 x 10^-6) = 6 x 10^-13.

delta P = dp/dLambda * deltaLambda

I'm not sure how to solve this relationship to get an expression for deltaLambda, to plug back into the original

deltaX > h/(4pi)(deltaP)

Once I get the expression for momentum in terms of lambda, I think I just sub it back in and solve for deltaX.

Any help? or have I done anything wrong so far?

2. Oct 14, 2007

### Astronuc

Staff Emeritus
Ok - One has $\Delta\lambda$, from which one can obtain $\Delta$E.

What is the relationship between E and p for a photon?

3. Oct 14, 2007

### zing478

Well, E = hc/$$\lambda$$

From de Broglies wavelength equation,
$$\lambda$$ = h/p

so here $$\Delta$$E=$$\Delta$$pc

So to find the uncertainty in position, I have $$\Delta$$$$\lambda$$, change that to $$\Delta$$E. From there I can get $$\Delta$$p, and

$$\Delta$$x = h/4$$\pi$$$$\Delta$$p

Where $$\Delta$$x is the minimum uncertainty in position?

Does that make sense?

4. Oct 14, 2007

### Astronuc

Staff Emeritus
Yes. That would do it.

5. Oct 14, 2007

### zing478

Thanks for the help!