The Heisenberg uncertainty principle is written Δx*mΔu ≥ h/(4π).
what is the minimum uncertainty in position (in Angstrom) of a hydrogen atom traveling at 1.05 * 10^4 m/s with an uncertainty of 0.10 * 10^4 m/s?
the answer is 0.31 Angstrom, but i dont' know how to get that. please explain!
what is the minimum uncertainty in position (in Angstrom) of a hydrogen atom traveling at 1.05 * 10^4 m/s with an uncertainty of 0.10 * 10^4 m/s?
the answer is 0.31 Angstrom, but i dont' know how to get that. please explain!
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Hey half the work has been done by the equation the uncertainty in position means you have to calculate Δx and for Δv take original -uncerianity value