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JEE Mains · Physics · STD 12 - 11. 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
- 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 \(\frac{\lambda_{0}}{\sqrt{{\Large{2}}+\cfrac{\mathrm{e}^{2} \mathrm{E}^{2} \mathrm{t}^{2}}{\mathrm{m}^{2} \mathrm{v}_{0}^{2}}}}\)
- C \(\frac{\lambda_{0}}{\sqrt{{\Large{1}}+\cfrac{\mathrm{e}^{2} \mathrm{E}^{2} \mathrm{t}^{2}}{2 \mathrm{m}^{2} \mathrm{v}_{0}^{2}}}}\)
- D \(\frac{\lambda_{0}}{\sqrt{{\Large{1}}+\cfrac{\mathrm{e}^{2} \mathrm{E}_{0}^{2} \mathrm{t}^{2}}{\mathrm{m}^{2} \mathrm{v}_{0}^{2}}}}\)
Answer & Solution
Correct Answer
(C) \(\frac{\lambda_{0}}{\sqrt{{\Large{1}}+\cfrac{\mathrm{e}^{2} \mathrm{E}^{2} \mathrm{t}^{2}}{2 \mathrm{m}^{2} \mathrm{v}_{0}^{2}}}}\)
Step-by-step Solution
Detailed explanation
By de-Broglie hypothesis \(\lambda=\frac{\mathrm{h}}{\mathrm{mv}}\) \(\lambda_{0}=\frac{\mathrm{h}}{\mathrm{m} \sqrt{2} \mathrm{v}_{0}}\dots(1)\)…
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