Measuring the wavelength and position of a photon

[zing478]

Homework Statement

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?

Homework Equations

deltaP*deltaX > h/4pi

p = h/lambda

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?

 

[Astronuc]

Ok - One has \Delta\lambda, from which one can obtain \DeltaE.

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

 

[zing478]

Well, E = hc/\lambda

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

so here \DeltaE=\Deltapc

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

\Deltax = h/4\pi\Deltap

Where \Deltax is the minimum uncertainty in position?

Does that make sense?

 

[Astronuc]

Yes. That would do it.

 

[zing478]

Thanks for the help!