CHEMISTRY COMMUNITY

Question: If the uncertainty in the momentum increases for a given particle what does this mean for the uncertainty in its wavelength?

I was wondering what the relationship between these two uncertainties are and how one can potentially affect the other?

Because of λ=h/p, an increase in uncertainty of p increases the uncertainty of λ.

I thought it would be the opposite? Because dividing planks constant by a larger number would lead to a smaller wavelength value? Because don't they have an indirect relationship?

The uncertainty in momentum is inversely proportional to the uncertainty in wavelength, meaning a larger uncertainty in momentum will lead to a smaller uncertainty in wavelength, and vice versa.

Because of the inverse relationship between momentum and wavelength shown in the de Broglie equation(λ=h/p), increasing the uncertainty of momentum will decrease the uncertainty of wavelength.

Because of the inverse relationship between momentum and wavelength shown in the de Broglie equation(λ=h/p), increasing the uncertainty of momentum will decrease the uncertainty of wavelength.

chrisavalos- 1K wrote:Question: If the uncertainty in the momentum increases for a given particle what does this mean for the uncertainty in its wavelength?

I was wondering what the relationship between these two uncertainties are and how one can potentially affect the other?

Because there's an inverse relationship between momentum and wavelength, an increase in the uncertainty of momentum will entail a decrease in the uncertainty of wavelength.

Because of the inverse relationship between momentum (p) and wavelength(lambda) is given in De Broglie's equation, lambda=h/p, therefore, when p is increased, the value of lambda decreases.