If the location of a particular electron can be measured only to a precision of 0.069 nm,what is the minimum uncertainty in the electron's velocity?
(m_e=9.109×10^-31 kg,c=3.00×10⁸ m/s,h=6.63×10^-34 J·s)

Question

Quizplus · Accepted Answer

The minimum uncertainty in the electron's velocity can be calculated using Heisenberg's uncertainty principle, which is given by Δx * Δv ≥ h / (4πm), where Δx is the uncertainty in position, Δv is the uncertainty in velocity, h is Planck's constant, and m is the mass of the electron. Substituting the given values (Δx = 0.069 nm = 0.069 × 10^-9 m, h = 6.63 × 10^-34 J·s, m = 9.109 × 10^-31 kg) into the formula and solving for Δv gives a minimum uncertainty in the electron's velocity of approximately 8.4 × 10^5 m/s.

If the Location of a Particular Electron Can Be Measured