Dual Nature of Matter

JEE Main level, single correct, medium

17 of 200

Question 17 of 200

An electron of mass ' m ' with an initial velocity v→=v0i^(v0>0)\overrightarrow{\mathrm{v}}=\mathrm{v}_0 \hat{i}\left(\mathrm{v}_0\gt0\right) enters an electric field E→=−Eok^\overrightarrow{\mathrm{E}}=-\mathrm{E}_{\mathrm{o}} \hat{\mathrm{k}}. If the initial de Broglie wavelength is λ0\lambda_0, the value after time t would be

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Answer: (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}}}

v→=v0i^−E0emtk^∣v→∣=v02+E02e2t2 m2λ0=hmv0λ′=hmv01+E02e2t2v02 m2λ′=λ01+E02e2t2v02 m2\begin{aligned} & \overrightarrow{\mathrm{v}}=\mathrm{v}_0 \hat{\mathrm{i}}-\frac{\mathrm{E}_0 \mathrm{e}}{\mathrm{m}} t \hat{\mathrm{k}} \\ & |\overrightarrow{\mathrm{v}}|=\sqrt{\mathrm{v}_0^2+\frac{\mathrm{E}_0^2 \mathrm{e}^2 \mathrm{t}^2}{\mathrm{~m}^2}} \\ & \lambda_0=\frac{\mathrm{h}}{\mathrm{mv}_0} \\ & \lambda^{\prime}=\frac{\mathrm{h}}{\mathrm{mv}_0 \sqrt{1+\frac{\mathrm{E}_0^2 \mathrm{e}^2 \mathrm{t}^2}{\mathrm{v}_0^2 \mathrm{~m}^2}}} \\ & \lambda^{\prime}=\frac{\lambda_0}{\sqrt{1+\frac{\mathrm{E}_0^2 \mathrm{e}^2 \mathrm{t}^2}{\mathrm{v}_0^2 \mathrm{~m}^2}}}\end{aligned}

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