JEE Mains · Physics · STD 12 - Dual nature of radiation and matter
- JEE Mains 2025
- Hard
- MCQ
An electron of mass ' m ' with an initial velocity \(\overrightarrow{\mathrm{v}}=\mathrm{v}_0 \hat{i}\left(\mathrm{v}_0\gt0\right)\) enters an electric field \(\overrightarrow{\mathrm{E}}=-\mathrm{E}_{\mathrm{o}} \hat{\mathrm{k}}\). If the initial de Broglie wavelength is \(\lambda_0\), the value after time t would be _______.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(\frac{\lambda_0}{\sqrt{1+\frac{\mathrm{e}^2 \mathrm{E}_0{ }^2 \mathrm{t}^2}{\mathrm{~m}^2 \mathrm{v}_0{ }^2}}}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \overrightarrow{\mathrm{v}}=\mathrm{v}_0 \hat{\mathrm{i}}-\frac{\mathrm{E}_0 \mathrm{e}}{\mathrm{m}} t \hat{\mathrm{k}} \\ & |\overrightarrow{\mathrm{v}}|=\sqrt{\mathrm{v}_0^2+\frac{\mathrm{E}_0^2 \mathrm{e}^2 \mathrm{t}^2}{\mathrm{~m}^2}} \\ & \lambda_0=\frac{\mathrm{h}}{\mathrm{mv}_0} \\ & \lambda^{\prime}=\frac{\mathrm{h}}{\mathrm{mv}_0 \sqrt{1+\frac{\mathrm{E}_0^2 \mathrm{e}^2 \mathrm{t}^2}{\mathrm{v}_0^2 \mathrm{~m}^2}}} \\ & \lambda^{\prime}=\frac{\lambda_0}{\sqrt{1+\frac{\mathrm{E}_0^2 \mathrm{e}^2 \mathrm{t}^2}{\mathrm{v}_0^2 \mathrm{~m}^2}}}\end{aligned}\)
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