An electron (mass m) with initial velocity v =v0 i +v0 j is in an electric field E =-E0 k. If λ0 is initial de-Broglie wavelength of electron, its de-Broglie wave length at time t is given by

(A) \frac{\lambda_{0} \sqrt{2}}{\sqrt{ {\Large{1}}+\cfrac{\mathrm{e}^{2} \mathrm{E}^{2} \mathrm{t}^{2}}{\mathrm{m}^{2} \mathrm{v}_{0}^{2}}}}

(B) \frac{\lambda_{0}}{\sqrt{{\Large{2}}+\cfrac{\mathrm{e}^{2} \mathrm{E}^{2} \mathrm{t}^{2}}{\mathrm{m}^{2} \mathrm{v}_{0}^{2}}}}

(D) \frac{\lambda_{0}}{\sqrt{{\Large{1}}+\cfrac{\mathrm{e}^{2} \mathrm{E}_{0}^{2} \mathrm{t}^{2}}{\mathrm{m}^{2} \mathrm{v}_{0}^{2}}}}

JEE Mains · Physics · STD 12 - Dual nature of radiation and matter

An electron (mass \(m\)) with initial velocity \(\overrightarrow{\mathrm{v}}=\mathrm{v}_{0} \hat{\mathrm{i}}+\mathrm{v}_{0} \hat{\mathrm{j}}\) is in an electric field \(\overrightarrow{\mathrm{E}}=-\mathrm{E}_{0} \hat{\mathrm{k}} .\) If \(\lambda_{0}\) is initial de-Broglie wavelength of electron, its de-Broglie wave length at time \(t\) is given by

Pick an option to check your answer.

Verified Solution

Correct Answer

(C) \(\frac{\lambda_{0}}{\sqrt{{\Large{1}}+\cfrac{\mathrm{e}^{2} \mathrm{E}^{2} \mathrm{t}^{2}}{2 \mathrm{m}^{2} \mathrm{v}_{0}^{2}}}}\)

Detailed explanation

By de-Broglie hypothesis

\(\lambda=\frac{\mathrm{h}}{\mathrm{mv}}\)

\(\lambda_{0}=\frac{\mathrm{h}}{\mathrm{m} \sqrt{2} \mathrm{v}_{0}}\dots(1)\)

\(\lambda^{\prime}=\frac{\mathrm{h}}{\sqrt{\mathrm{v}_{0}^{2}+\mathrm{v}_{0}^{2}+\left(\frac{\mathrm{eE}_{0} \mathrm{t}}{\mathrm{m}}\right)^{2}}}\)

\(=\frac{h}{m \sqrt{2 v_{0}^{2}+\frac{e^{2} E_{0}^{2} t^{2}}{m^{2}}}}\dots(2)\)

\(\mathrm{By}(1)\) and ( 2)

\(\lambda^{\prime}=\frac{\lambda_{0}}{\sqrt{1+\frac{\mathrm{e}^{2} \mathrm{E}_{0}^{2} \mathrm{t}^{2}}{2 \mathrm{m}^{2} \mathrm{v}_{0}^{2}}}}\)

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