JEE MAIN - Physics (2022 - 27th July Morning Shift - No. 15)
An electron (mass $$\mathrm{m}$$) with an initial velocity $$\vec{v}=v_{0} \hat{i}\left(v_{0}>0\right)$$ is moving in an electric field $$\vec{E}=-E_{0} \hat{i}\left(E_{0}>0\right)$$ where $$E_{0}$$ is constant. If at $$\mathrm{t}=0$$ de Broglie wavelength is $$\lambda_{0}=\frac{h}{m v_{0}}$$, then its de Broglie wavelength after time t is given by
$$\lambda_{0}$$
$$\lambda_{0}\left(1+\frac{e E_{0} t}{m v_{0}}\right)$$
$$\lambda_{0} t$$
$$\frac{\lambda_{0}}{\left(1+\frac{e E_{0} t}{m v_{0}}\right)}$$
Explanation
$$
\text { At } t=0, \lambda_0=\frac{h}{m v_0}
$$
Since $\vec{v}=v_0 \hat{i}$ and $\overrightarrow{\mathrm{E}}=\mathrm{E}_0 \hat{i}$
its velocity $v$ at any time $t$ is given by
$$ v=v_o+\frac{\varepsilon \mathrm{E}_{\mathrm{o}}}{m} t $$
De Broglie wavelength $\lambda$ at any time $t$ is given by
$$ \begin{aligned} \lambda & =\frac{h}{m v}=\frac{h}{m\left(v_0+\frac{e \mathrm{E}_0}{m} t\right)} \\\\ & =\frac{h}{m v_0\left(1+\frac{e \mathrm{E}_0}{m v_0} t\right)} \\\\ & =\frac{\lambda_0}{1+\frac{e \mathrm{E}_0}{m v_0} t} \end{aligned} $$
Since $\vec{v}=v_0 \hat{i}$ and $\overrightarrow{\mathrm{E}}=\mathrm{E}_0 \hat{i}$
its velocity $v$ at any time $t$ is given by
$$ v=v_o+\frac{\varepsilon \mathrm{E}_{\mathrm{o}}}{m} t $$
De Broglie wavelength $\lambda$ at any time $t$ is given by
$$ \begin{aligned} \lambda & =\frac{h}{m v}=\frac{h}{m\left(v_0+\frac{e \mathrm{E}_0}{m} t\right)} \\\\ & =\frac{h}{m v_0\left(1+\frac{e \mathrm{E}_0}{m v_0} t\right)} \\\\ & =\frac{\lambda_0}{1+\frac{e \mathrm{E}_0}{m v_0} t} \end{aligned} $$
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