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Physics · Atomic Physics

JEE Main 2025 — 24 January, Morning Shift — Question 63

An electron of mass ' mm ' with an initial velocity v→=v0i^(v0>0)\overrightarrow{\mathrm{v}}=\mathrm{v}_{0} \hat{\mathrm{i}}\left(\mathrm{v}_{0}>0\right) \quad enters an electric field E→=−E0k^\overrightarrow{\mathrm{E}}=-\mathrm{E}_{0} \hat{\mathrm{k}}. If the initial de Broglie wavelength is λ0\lambda_{0}, the value after time tt would be :-

  1. Option A:

    λ01+e2E02t2 m2v02\frac{\lambda_{0}}{\sqrt{1+\frac{\mathrm{e}^{2} \mathrm{E}_{0}^{2} \mathrm{t}^{2}}{\mathrm{~m}^{2} \mathrm{v}_{0}^{2}}}}

    Correct
  2. Option B:

    λ01−e2E02t2m2v02\frac{\lambda_{0}}{\sqrt{1-\frac{e^{2} E_{0}^{2} t^{2}}{m^{2} v_{0}^{2}}}}

  3. Option C:

    λ0\lambda_{0}

  4. Option D:

    λ01+e2E02t2m2v02\lambda_{0} \sqrt{1+\frac{e^{2} E_{0}^{2} t^{2}}{m^{2} v_{0}^{2}}}

Answer: A

Step-by-step solution

v→=v0i^−E0emtk^\overrightarrow{\mathrm{v}}=\mathrm{v}_{0} \hat{\mathrm{i}}-\frac{\mathrm{E}_{0} \mathrm{e}}{\mathrm{m}} t \hat{\mathrm{k}}

∣v→∣=v02+E02e2t2 m2|\overrightarrow{\mathrm{v}}|=\sqrt{\mathrm{v}_{0}^{2}+\frac{\mathrm{E}_{0}^{2} \mathrm{e}^{2} \mathrm{t}^{2}}{\mathrm{~m}^{2}}}

λ0=hmv0\lambda_{0}=\frac{\mathrm{h}}{\mathrm{mv}_{0}}

λ′=hmv⁡01+E02e2t2v02 m2\lambda^{\prime}=\frac{h}{\operatorname{mv}_{0} \sqrt{1+\frac{\mathrm{E}_{0}^{2} \mathrm{e}^{2} \mathrm{t}^{2}}{\mathrm{v}_{0}^{2} \mathrm{~m}^{2}}}}

λ′=λ01+Ee2e2t2v02 m2\lambda^{\prime}=\frac{\lambda_{0}}{\sqrt{1+\frac{\mathrm{E}_{\mathrm{e}}^{2} \mathrm{e}^{2} \mathrm{t}^{2}}{\mathrm{v}_{0}^{2} \mathrm{~m}^{2}}}}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
24 January, Morning Shift
Subject
Physics
Chapter
Atomic Physics
Topic
Dual Nature of Matter

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