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Difficulty: MediumWave-Particle Duality and de Broglie Wavelength

A subatomic particle of mass 6.63×10−31 kg6.63 \times 10^{-31}\text{ kg} moves with a velocity of 1.0×106 m/s1.0 \times 10^6\text{ m/s}. Calculate its de Broglie wavelength in nanometers (nm\text{nm}). (Take Planck's constant h=6.63×10−34 J⋅sh = 6.63 \times 10^{-34}\text{ J}\cdot\text{s} and 1 nm=10−9 m1\text{ nm} = 10^{-9}\text{ m})

Answer: 1 nm

Answer

The de Broglie wavelength of the particle is 1.0 nm1.0\text{ nm}.
Using de Broglie's wave-particle duality relation λ=hp=hmv\lambda = \frac{h}{p} = \frac{h}{m v}, substituting the given values yields λ=6.63×10−34 J⋅s(6.63×10−31 kg)×(1.0×106 m/s)=1.0×10−9 m\lambda = \frac{6.63 \times 10^{-34}\text{ J}\cdot\text{s}}{(6.63 \times 10^{-31}\text{ kg}) \times (1.0 \times 10^6\text{ m/s})} = 1.0 \times 10^{-9}\text{ m}. In nanometers, this is equal to 1.0 nm1.0\text{ nm}.

Step-by-Step Solution

1
Calculate the linear momentum of the particle
p=6.63×10−25 kg⋅m/sp = 6.63 \times 10^{-25}\text{ kg}\cdot\text{m/s}
Momentum is the product of mass and velocity (p=mvp = m v).
2
Calculate the de Broglie wavelength
λ=1.0×10−9 m\lambda = 1.0 \times 10^{-9}\text{ m}
According to de Broglie's hypothesis, wavelength is given by λ=hp\lambda = \frac{h}{p}.
3
Convert the calculated wavelength to nanometers
λ=1.0 nm\lambda = 1.0\text{ nm}
Since 1 nm=10−9 m1\text{ nm} = 10^{-9}\text{ m}, dividing 1.0×10−9 m1.0 \times 10^{-9}\text{ m} by 10−910^{-9} yields 1.0 nm1.0\text{ nm}.

Key Concept

de Broglie Wavelength and Wave-Particle Duality
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Topics

Wave-Particle Duality and de Broglie Wavelength

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