Heisenberg Uncertainy Principle

AI Thread Summary
The discussion focuses on the calculations related to the Heisenberg Uncertainty Principle and the de Broglie wavelength of electrons in a beam emitted by modern electron microscopes. The wavelength of the electrons is calculated to be approximately 4.8494 x 10^-12 m. A minimum uncertainty in electron position of 1 angstrom (10^-10 m) leads to a maximum acceptable uncertainty in momentum calculated as 5.27 x 10^-25 kg·m/s. Participants clarify that the wavelength calculation pertains to de Broglie rather than the uncertainty principle itself. The conversation emphasizes the importance of distinguishing between these two concepts in quantum mechanics.
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Homework Statement


Modern electron microscopes used in biological research emit a beam of electrons with a
velocity of 1.5x108 m/s
a. What is the wavelength of an electron in the beam?
b. The wavelength of the particle determines the resolution of the microscopy. Assume
that you desire a minimum uncertainty in electron position of 1A. Using the uncertainty
principle, what is the maximum acceptable uncertainty in the momentum of the
electrons?

Homework Equations



lamda = h/mv
p_0 = h/(4pi*delta x)

The Attempt at a Solution



a) lamda = h/mv = 4.8494*10^-12m

b) p_0 = h/(4pi*delta x) = 5.27*10^-25 kg*ms^-1? thanks
 
Physics news on Phys.org
a) is not Heisenbergs uncertainty principle, this is de broglie wavelength

b) is "1A" one "angstrom" 10-10m?
 

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