Heisenberg Uncertainy Principle

[Dars]

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

 

[malawi_glenn]

a) is not Heisenbergs uncertainty principle, this is de broglie wavelength

b) is "1A" one "angstrom" 10^-10m?