Final answer:
Using the Heisenberg uncertainty principle, we can calculate that the minimum uncertainty in the speed of an electron confined to the diameter of an atom (approximately 100 pm) is about 5.8 × 105 m/s.
Step-by-step explanation:
To calculate the minimum uncertainty in the speed of an electron confined to a region with the diameter of an atom, approximately 100 pm, we can use the Heisenberg uncertainty principle. This principle is typically stated as ∆x∆p ≥ ħ/2, where ∆x is the uncertainty in position, ∆p is the uncertainty in momentum, and ħ (h-bar) is the reduced Planck's constant, which has a value of approximately 1.055 × 10-34 kg m2/s. Given the uncertainty in position (∆x) to be 100 pm (1 × 10-10 m), the minimum uncertainty in momentum (∆p) can be calculated:
∆p = ħ / (2 × ∆x)
∆p = 1.055 × 10-34 kg m2/s / (2 × 1 × 10-10 m)
∆p ≈ 5.275 × 10-25 kg m/s
To find the uncertainty in speed (∆v), we divide the uncertainty in momentum by the mass of the electron (me), approximately 9.11 × 10-31 kg:
∆v = ∆p / me
∆v ≈ (5.275 × 10-25 kg m/s) / (9.11 × 10-31 kg)
∆v ≈ 5.79 × 105 m/s
The minimum uncertainty in the speed of the electron is therefore approximately 5.8 × 105 m/s, reported to two significant figures.