Answer:
His de Broglie wavelength is 8.35×10⁻³⁷ m
The uncertainty in his position is 1.15 × 10⁻³⁵ m
Step-by-step explanation:
First, Convert 227-lb to kg and convert the unit of the speed from mi/h to m/s.
To convert 227-lb to kg,
1-lb = 0.453592 kg
∴ 227-lb = 227 × 0.453592 kg
227-lb = 102.97 kg
To convert 17.25 ± 0.10 mi/h to m/s
1 mi = 1609.34 m
and 1 h = 3600 s
Therefore,
17.25 mi/h = (17.25 ×1609.34)/3600 m/s = 7.71 m/s
and 0.10 mi/h = (0.10 ×1609.34)/3600 m/s = 0.044704 m/s
Hence, the speed 17.25 ± 0.10 mi/h = 7.71 ± 0.044704 m/s
Now
(a) To determine the de Broglie wavelength,
De Broglie wavelength is given by
λ = h/mv
Where λ is the de Broglie wavelength
h is Planck's constant (h = 6.626×10⁻³⁴ kgm²/s)
m is the mass
and v is the speed (velocity)
From the question
m = 102.97 kg
v = 7.71 m/s
Therefore,
λ = 6.626×10⁻³⁴ / (102.97×7.71)
λ = 8.35×10⁻³⁷ m
Hence, his de Broglie wavelength is 8.35×10⁻³⁷ m
(b) To calculate the uncertainty in his position
From
Δx = h/(4πmΔv)
Where Δx is the uncertainty in the position
h is Planck's constant (h = 6.626×10⁻³⁴ kgm²/s)
π is a constant ( π = 3.14)
m is the mass
Δv is the uncertainty in speed
From the question
m = 102.97 kg
Δv = 0.044704 m/s
Hence,
Δx = 6.626×10⁻³⁴ / (4×3.14×102.97×0.044704)
Δx = 1.15 × 10⁻³⁵ m
Hence, the uncertainty in his position is 1.15 × 10⁻³⁵ m.