Question
An electron of mass m is moving in an electric field $\overrightarrow{E} = - 2E_{0}\widehat{i}\left( E_{0} = \right.\ $ constant $\left. \ > 0 \right)$, with an initial velocity $\overrightarrow{V} = v_{o}\widehat{i}\ \left( v_{0} = \right.\ $ constant $\left. \ > 0 \right)$. If $\lambda_{0} = \frac{h}{4{mv}_{0}}$, its de Broglie wavelength at time t is
$\_\_\_\_$ . ( $e =$ charge of electron)
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Solution
Sol. [Key Concept: $v = v_{0} + at$ with $a = \frac{2{eE}_{0}}{m}$; then $\lambda = \frac{h}{mv} = \frac{4\lambda_{0}}{1 + \frac{2E_{0}e}{m}\frac{t}{v_{0}}}$.]
$$V = U + at$$
$$\begin{matrix} & V = V_{0} + \left( \frac{2{eE}_{0}}{\text{ }m} \right)t \\ & \ \because a = \frac{\left( 2E_{0} \right)e}{\text{ }m} \\ & \lambda = \frac{h}{mV} \\ & \lambda = \frac{h}{\text{ }m\left\lbrack {\text{ }V}_{0} + \frac{2{eE}_{0}t}{\text{ }m} \right\rbrack} \end{matrix}$$
Put $\lambda_{0} = \frac{h}{4{mV}_{0}} \Rightarrow \text{ }h = 4\lambda_{0}{mV}_{0}$
$$\begin{matrix} \lambda & \ = \frac{4\text{ }m\lambda_{0}{\text{ }V}_{0}}{\text{ }m\left\lbrack {\text{ }V}_{0} + \frac{2{eE}_{0}}{\text{ }m}t \right\rbrack} \\ \lambda & \ = \frac{4\lambda_{0}}{1 + \frac{2{eE}_{0}}{{mV}_{0}}t} \end{matrix}$$
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