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Difficulty: MediumWave-Particle Duality and de Broglie Wavelength

A particle of constant mass has a de Broglie wavelength of λ\lambda. If the kinetic energy of the particle is increased to nine times its initial value, what is its new de Broglie wavelength?

  1. A
    λ9\frac{\lambda}{9}
  2. λ3\frac{\lambda}{3}Answer
  3. C
    3λ3\lambda
  4. D
    9λ9\lambda

Answer

The new de Broglie wavelength is λ3\frac{\lambda}{3}.
The de Broglie wavelength of a particle is given by λ=h2mEk\lambda = \frac{h}{\sqrt{2mE_k}}, showing that wavelength is inversely proportional to the square root of its kinetic energy. When kinetic energy is increased by a factor of 9, the square root factor becomes 9=3\sqrt{9} = 3, reducing the new wavelength to one-third of its original value (λ3\frac{\lambda}{3}).

Step-by-Step Solution

1
Relate de Broglie wavelength to momentum and kinetic energy.
λ=hp=h2mEk\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE_k}}
Since kinetic energy is Ek=p22mE_k = \frac{p^2}{2m}, linear momentum is p=2mEkp = \sqrt{2mE_k}.
2
Set up the ratio for the new kinetic energy Ek′=9EkE_k' = 9E_k.
λ′=h2m(9Ek)=h32mEk\lambda' = \frac{h}{\sqrt{2m(9E_k)}} = \frac{h}{3\sqrt{2mE_k}}
Substitute the new kinetic energy into the de Broglie wavelength equation.
3
Express the new wavelength in terms of the initial wavelength λ\lambda.
λ′=λ3\lambda' = \frac{\lambda}{3}
Comparing λ′\lambda' to λ=h2mEk\lambda = \frac{h}{\sqrt{2mE_k}} gives λ′=λ3\lambda' = \frac{\lambda}{3}.

Key Concept

De Broglie Wavelength and Kinetic Energy Relationship
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Topics

Wave-Particle Duality and de Broglie Wavelength

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