Uncertainty in wavelength and position

[Saptarshi Sarkar]

Homework Statement

Wavelengths can be measured with accuracies of one part in 10⁴. What is the uncertainty in the position of a 1A X-Ray photon when its wavelength is simultaneously measured?

Relevant Equations

$ΔxΔp>=\frac {\hbar} 2$

In my attempt to solve the problem, I used the formula for de-Broglie wavelength $p=\frac h λ$ and differentiated both sides to get $Δp = \frac {hΔλ} {λ^2}$.

Plugging this equation into the Heisenberg's position and momentum uncertainty principle formula and calculating the minimum uncertainty in momentum, I got

$Δx=\frac {\hbarλ^2} {2hΔλ} = 7.96 × 10^{-22} m$

The solution looks toο small to me. I am not sure if the value of $Δλ$ is $10^{-4}$ which I used in the calculation. I am not quite sure what one part in $10^4$ means for the uncertainty in λ.

 

[kuruman]

You are given that $\dfrac{\Delta \lambda}{\lambda}=\dfrac{1}{10^4}$. Did you use that in your equation for $\Delta x$?

 

[Saptarshi Sarkar]

You are given that $\dfrac{\Delta \lambda}{\lambda}=\dfrac{1}{10^4}$. Did you use that in your equation for $\Delta x$?

No, I used $Δλ = 10^{-4}$. I was confused what accuracy of one part in $10^4$ meant.

The answer should be $7.96 \times 10^{-12}m = 0.0796 A$ then.

Thanks for the help!

 

[PeroK]

Plugging this equation into the Heisenberg's position and momentum uncertainty principle formula and calculating the minimum uncertainty in momentum, I got

$Δx=\frac {\hbarλ^2} {2hΔλ} = 7.96 × 10^{-22} m$

The solution looks toο small to me. I am not sure if the value of $Δλ$ is $10^{-4}$ which I used in the calculation. I am not quite sure what one part in $10^4$ means for the uncertainty in λ.

I think you may have taken $\Delta \lambda = 10^4m$ there!

"One part in $10^4$" means $0.01 \%$

 

[kuruman]

No, I used $Δλ = 10^{-4}$. I was confused what accuracy of one part in $10^4$ meant.

The answer should be $7.96 \times 10^{-12}m = 0.0796 A$ then.

Thanks for the help!

Considering that this is an order of magnitude calculation, I would go with 0.08 A to one significant figure.

 

[Cutter Ketch]

Is it just me, or is anyone else getting $10^4$ larger?

 

[PeroK]

Is it just me, or is anyone else getting $10^4$ larger?

I agree, $\lambda^2 = 10^{-20}m, \ \Delta \lambda = 10^{-14}m$, which leads to $\approx \frac{1}{12} \times 10^{-6}m \approx 8 \times 10^{-8}m$

 

[Cutter Ketch]

I think this is just a calculator error. It helps if you remember what @kuruman said in post 2 and simplify your expression for $\Delta$x algebraicly before putting in numbers.