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Quantum Physics: Question 6

Syllabus 22.3

Multiple choice A2 1 mark

In an electron gun, an electron travels with speed v=2.00×106 m s−1v=2.00\times10^{6}\text{ m s}^{-1}.

What is the de Broglie wavelength of this electron? (mass of electron =9.11×10−31 kg=9.11\times10^{-31}\text{ kg})

Choose an answer to check it, then compare with the worked solution below.
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Worked solution

Step 1: Calculate the momentum of the electron

The de Broglie wavelength of a moving particle is: λ=hp=hmv\lambda = \frac{h}{p} = \frac{h}{mv}

First find the momentum p=mvp=mv, using m=9.11×10−31 kgm=9.11\times10^{-31}\text{ kg} and v=2.00×106 m s−1v=2.00\times10^{6}\text{ m s}^{-1}: p=9.11×10−31×2.00×106=1.822×10−24 kg m s−1p = 9.11\times10^{-31}\times2.00\times10^{6} = 1.822\times10^{-24}\text{ kg m s}^{-1}

Step 2: Apply the de Broglie relation

Substituting h=6.63×10−34 J sh=6.63\times10^{-34}\text{ J s}: λ=6.63×10−341.822×10−24=3.64×10−10 m\lambda = \frac{6.63\times10^{-34}}{1.822\times10^{-24}} = 3.64\times10^{-10}\text{ m}

Recompute as a check: 3.64×10−10×1.822×10−24=6.63×10−34 J s3.64\times10^{-10}\times1.822\times10^{-24}=6.63\times10^{-34}\text{ J s}, which matches hh. Consistent.

Why the other options are wrong

  • B (3.64×10−7 m3.64\times10^{-7}\text{ m}): comes from confusing the exponent of the electron’s mass with that of the Planck constant, using m=9.11×10−34 kgm=9.11\times10^{-34}\text{ kg} instead of 9.11×10−31 kg9.11\times10^{-31}\text{ kg}; this shrinks the momentum, and hence inflates λ\lambda, by a factor of 10001000.
  • C (1.99×10−13 m1.99\times10^{-13}\text{ m}): comes from using the mass of a proton, mp=1.67×10−27 kgm_p=1.67\times10^{-27}\text{ kg}, instead of the mass of an electron.
  • D (1.46×103 m1.46\times10^{3}\text{ m}): comes from inverting the relation and calculating λ=hvm\lambda=\dfrac{hv}{m} instead of λ=hmv\lambda=\dfrac{h}{mv}.

Final answer

  • The de Broglie wavelength is 3.64×10−10 m\boxed{3.64\times10^{-10}}\text{ m}, option A.