An electron has a kinetic energy of 1.5 eV.If its momentum is uncertain by ± 2.5%,the minimum uncertainty in its position is approximately
A)3.2 nm
B)5.0 nm
C)1.0 nm
D)2.2 nm
E)1.6 nm

Question

Quizplus · Accepted Answer

The uncertainty in position is given by the Heisenberg uncertainty principle: ΔxΔp ≥ h/4π, where Δx is the uncertainty in position, Δp is the uncertainty in momentum, and h is Planck's constant. We are given that the uncertainty in momentum is ± 2.5%, so Δp = 0.025p. We can use the kinetic energy of the electron to find its momentum: p = √(2mK), where m is the mass of the electron and K is its kinetic energy. Plugging in the values for m and K, we get p = 5.4 × 10^-25 kg m/s. Substituting this into the uncertainty principle, we get Δx ≥ 3.2 nm.

An Electron Has a Kinetic Energy of 1