Question 21 of Instructional Module: Heisenberg Uncertainty Principle

[MiaRosati3L]

Question 21 asks: The electron is not confined to the nucleus and we now know that the size of an atom is determined by its electrons outside of the nucleus. For hydrogen its measured atomic diameter is 145,000 times its nuclear diameter of 1.7 x 10-15 m. In other words make the (correct) assumption that the electron is confined to the atomic diameter of 2.5 x 10-10 m which would be a better estimate of the electron's uncertainty in position. Repeat all of the above calculations and comment on your values. Use the Heisenberg uncertainty equation to calculate the electron's uncertainty in momentum. Then use the mass of electron 9.1 x 10-31 kg to calculate its uncertainty in velocity. Comment on your value.

My brain is fatigued and numbers no longer have meaning. Could someone please walk me through this problem?

[Ashley M 2E]

Basically, the question is asking you to repeat what you did for question 19 but instead use the correct uncertainty of the atomic diameter as your uncertainty in position. Using that value, find the uncertainty in the velocity.

[Gabrielle Malte 2G]

The correct model for interpreting uncertainty in an electron's position and momentum is to use the atomic diameter (1.7x10^-15m), as electrons are confined to the space of the whole atom rather than just the nucleus. Since we don't know the exact location of the electron in the atom, the uncertainty in position, delta x, is 1.7X10^-15m. To find uncertainty in velocity, we use the Heisenberg uncertainty equation and set delta v > or = h/m4pi(1.7x10^-15m).
Hope this helps!