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JEE Main 2020 (8 Jan, Shift 2) · Matter Waves: de Broglie Wavelength

Matter Waves: de Broglie Wavelength question from JEE Main 2020 (8 Jan, Shift 2)

Q117JEE Main 2020 · 8 JAN · SHIFT 2Medium

Matter Waves: de Broglie Wavelength

An electron (mass m) with initial velocity v⃗=v0i^+v0j^\vec{v} = v_0\hat{i} + v_0\hat{j} is in an electric field E⃗=−E0k^\vec{E} = -E_0\hat{k}. If λ0\lambda_0 is initial de-Broglie wavelength of electron, its de-Broglie wave length at time t is given by:

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Answer: (c)

  1. Initial speed 2v0\sqrt2 v_0, so λ0=h2mv0\lambda_0 = \dfrac{h}{\sqrt2 mv_0}.
  2. Force eE0k^eE_0\hat k adds vz=eE0tmv_z = \dfrac{eE_0t}{m}; v2=2v02+e2E02t2m2v^2 = 2v_0^2 + \dfrac{e^2E_0^2t^2}{m^2}.
  3. λ=hmv=λ01+e2E02t22m2v02\lambda = \dfrac{h}{mv} = \dfrac{\lambda_0}{\sqrt{1+\frac{e^2E_0^2t^2}{2m^2v_0^2}}}
  4. So the answer is (c).
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