Determine the De Broglie’s wavelength of a car (m = 3.00 x 10 3 lb) traveling at 55.0 mi.hour -1 . Data: 1 kg = 2.205 lb and 1 km = 0.6214 mi. | General Chemistry 1

λ = h / p = h / mv m = 3 . 00 x 10 3 lb × 1 kg 2 . 205 lb = 1.36 x 10 3 kg v = 55 mi 1 hour × 1000 m 0 . 6214 mi × 1 hour 3600 s = 24.6 m.s -1 λ = 6 . 626 × 10 - 34 1 . 36 × 10 3 × 24 . 6 = 1.98 x 10 -35 m

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Wave-particle duality - 2 | Early Quantum Theory

Chapter 4 - Early Quantum Theory

Determine the De Broglie’s wavelength of a car (m = 3.00 x 10³ lb) traveling at 55.0 mi.hour^-1. Data: 1 kg = 2.205 lb and 1 km = 0.6214 mi.

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Solution

λ = h / p = h / mv m =

\frac{(3.00 x 10^{3} lb) \times (1 kg)}{2.205 lb}

= 1.36 x 10³ kg

v = $\frac{55 mi}{1 hour} \times \frac{1000 m}{0.6214 mi} \times \frac{1 hour}{3600 s}$ = 24.6 m.s^-1

λ = $\frac{6.626 \times 10^{- 34}}{1.36 \times 10^{3} \times 24.6}$ = 1.98 x 10^-35 m

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Wave-particle duality - 2 | Early Quantum Theory

General Chemistry 1 - Wave-particle duality - 2