An electron of mass 'm' with an initial velocity v→=v0i^(v0>0) enters an electric field E→=−E0k^.. If the initial de Broglie wavelength is λ0, the value after time t would be :-

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An electron of mass 'm' with an initial velocity $\vec{v} = v_{0} \hat{i} (v_{0} > 0)$ enters an electric field $\vec{E} = - E_{0} \hat{k} .$ . If the initial de Broglie wavelength is $λ_{0}$ , the value after time t would be :-

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$\frac{λ_{0}}{\sqrt{1 - \frac{e^{2} E_{0}^{2} t^{2}}{m^{2} v_{0}^{2}}}}$

$λ_{0} \sqrt{1 + \frac{e^{2} E_{0}^{2} t^{2}}{m^{2} v_{0}^{2}}}$

$λ_{0}$

$\frac{λ_{0}}{\sqrt{1 + \frac{e^{2} E_{0}^{2} t^{2}}{m^{2} v_{0}^{2}}}}$

answer is A.

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Detailed Solution

$\begin{aligned} \vec{v} = v_{0} \hat{i} - \frac{E_{0} e}{m} t \hat{k} \\ | \vec{v} | = \sqrt{v_{0}^{2} + \frac{E_{0}^{2} e^{2} t^{2}}{m^{2}}} \\ λ_{0} = \frac{h}{mv} \\ λ^{'} = \frac{h}{{mv}_{0} \sqrt{1 + \frac{E_{0}^{2} e^{2} t^{2}}{v_{0}^{2} m^{2}}}} \\ λ^{'} = \frac{λ_{0}}{\sqrt{1 + \frac{E_{0}^{2} e^{2} t^{2}}{v_{0}^{2} m^{2}}}} \end{aligned}$