An electron of charge $e$ and mass $m$ moving with an initial velocity $v_0 \hat{\mathbf{i}}$ is subjected…

An electron of charge $e$ and mass $m$ moving with an initial velocity $v_0 \hat{\mathbf{i}}$ is subjected to all electric field $E_0 \hat{\mathbf{j}}$. The de-Broglie wavelength of the electron at a time $t$ is (Initial de-Broglie wavelength of the electron $\left.=\lambda_0\right)$
  1. $\lambda_0$
  2. $\lambda_0 \sqrt{1+\frac{e^2 E_0^2 t^2}{m^2 v_0^2}}$
  3. $\frac{\lambda_0}{\sqrt{1+\frac{e^2 E_0^2 t^2}{m^2 v_0^2}}}$
  4. $\frac{\lambda_0}{\left(1+\frac{e^2 E_0^2 t^2}{m v_0^2}\right)}$

Solution

$\because$ de-Broglie relation of a charged particles, $ \lambda=\frac{h}{m v} $ Velocity of charged particle at time $t$, $ \mathbf{v}=v_0 \hat{\mathbf{i}}+\frac{e E_0}{m} t \hat{\mathbf{j}} \text { or }|v|=\sqrt{v_0^2+\left(\frac{e E_0}{m} t\right)^2} $ Hence, $\lambda=\frac{h}{m \sqrt{v_0^2+\left(\frac{e E_0}{m} t\right)^2}}$ or $ \begin{gathered} \lambda=\frac{h}{m v_0 \sqrt{1+\frac{e^2 E_0^2 t^2}{m^2 v_0^2}}} \text { or } \lambda=\frac{\lambda_0}{\sqrt{1+\frac{e^2 E_0^2 t^2}{m^2 v_0^2}}} \\ \left(\because \lambda_0=\frac{h}{m v_0}\right) \end{gathered} $

Asked in: AP EAMCET 2019 (20 Apr Shift 2)

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