Question

Suppose the velocity of an electron in an atom is known to an accuracy of 2.0×10^3 m/s (reasonably accurate compared with orbital velocities). What is the electron’s minimum uncertainty in position, and how does this compare with the approximate 0.1-nm size of the atom?

a) 7.35 x 10^(-11) m, Uncertainty is much smaller
b) 1.26 x 10^(-10) m, Uncertainty is comparable
c) 2.18 x 10^(-10) m, Uncertainty is much larger
d) 3.04 x 10^(-10) m, Uncertainty is much larger

Answer 1

Using Heisenberg's Uncertainty Principle, the minimum uncertainty in an electron's position within an atom is on the order of 10^-10 meters, which is comparable to the atom's size, thereby impacting our ability to accurately determine the electron's location.

Step-by-step explanation:

The question deals with the minimum uncertainty in the position of an electron in an atom when given the uncertainty in its velocity, using Heisenberg’s Uncertainty Principle. We are asked to compare this uncertainty with the size of the atom, which is approximately 0.1 nm. The Heisenberg Uncertainty Principle states that the product of the uncertainties in position and momentum of a particle cannot be smaller than a certain value, which is Planck’s constant divided by 4π. Since momentum is the product of mass and velocity, we can rewrite the uncertainty relation as Δx * m * Δ5v ≥ ℏ/(4π). The mass (m) of an electron is 9.11×10⁻³¹ kg, and ℏ (reduced Planck’s constant) is approximately 1.054571800×10⁻³´ J·s. Given the uncertainty in velocity (Δ7v) as 2.0×10³ m/s, we can calculate the uncertainty in position (Δx).

The calculation shows that the minimum uncertainty in position is on the order of 10⁻ⁱ⁰ meters, which is comparable in magnitude to the size of the atom. Therefore, we can deduce that the uncertainty in the position of an electron within an atom is significant compared to the size of the atom itself, affecting our ability to precisely determine its location while maintaining an accurate measure of its velocity.