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Difficulty: MediumWave-Particle Duality and de Broglie Wavelength

A subatomic particle of mass 2.0×10−27 kg2.0 \times 10^{-27}\text{ kg} possesses a kinetic energy of 1.0×10−19 J1.0 \times 10^{-19}\text{ J}. Given that Planck's constant h=6.6×10−34 J sh = 6.6 \times 10^{-34}\text{ J s}, what is the de Broglie wavelength of the particle?

  1. 3.3×10−11 m3.3 \times 10^{-11}\text{ m}Answer
  2. B
    4.7×10−11 m4.7 \times 10^{-11}\text{ m}
  3. C
    1.65×10−11 m1.65 \times 10^{-11}\text{ m}
  4. D
    3.3×10−57 m3.3 \times 10^{-57}\text{ m}

Answer

3.3×10−11 m3.3 \times 10^{-11}\text{ m}
The momentum of the particle is given by p=2mEk=2×2.0×10−27×1.0×10−19=2.0×10−23 kg m/sp = \sqrt{2mE_k} = \sqrt{2 \times 2.0 \times 10^{-27} \times 1.0 \times 10^{-19}} = 2.0 \times 10^{-23}\text{ kg m/s}. Substituting into the de Broglie wavelength relation λ=hp\lambda = \frac{h}{p} gives λ=6.6×10−342.0×10−23=3.3×10−11 m\lambda = \frac{6.6 \times 10^{-34}}{2.0 \times 10^{-23}} = 3.3 \times 10^{-11}\text{ m}.

Step-by-Step Solution

1
Relate kinetic energy to momentum
p=2mEk=2×(2.0×10−27 kg)×(1.0×10−19 J)=4.0×10−46=2.0×10−23 kg m/sp = \sqrt{2mE_k} = \sqrt{2 \times (2.0 \times 10^{-27}\text{ kg}) \times (1.0 \times 10^{-19}\text{ J})} = \sqrt{4.0 \times 10^{-46}} = 2.0 \times 10^{-23}\text{ kg m/s}
The de Broglie wavelength formula requires linear momentum, which connects to kinetic energy via Ek=p22mE_k = \frac{p^2}{2m}.
2
Calculate the de Broglie wavelength
λ=hp=6.6×10−34 J s2.0×10−23 kg m/s=3.3×10−11 m\lambda = \frac{h}{p} = \frac{6.6 \times 10^{-34}\text{ J s}}{2.0 \times 10^{-23}\text{ kg m/s}} = 3.3 \times 10^{-11}\text{ m}
Applying de Broglie's wave-particle duality relation λ=hp\lambda = \frac{h}{p}.

Key Concept

de Broglie Wavelength and Kinetic Energy Relation
Estimated Time:1m 30s
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Topics

Wave-Particle Duality and de Broglie Wavelength

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