An electron of mass m with an initial velocity V → = V 0 i ^ ( V 0 > 0 ) enters in an electric field E →…

1. ${\lambda }_0 t$
2. ${\lambda }_0\left(1+\frac{e{E}_0}{m{V}_0}t\right)$
3. $\frac{{\lambda }_0}{\left(1+\frac{e{E}_0}{m{V}_0}t\right)}$
4. ${\lambda }_0$

Solution

$\overrightarrow{V}=V_0\hat{\mathrm{i}}$
$\overrightarrow{E}=-E_0\hat{\mathrm{i}}$
$\overrightarrow{a}=\frac{\overrightarrow{F}}{m}=\frac{q\overrightarrow{E}}{m}=\frac{\left(-e\right)\left(-E_0\right)\hat{\mathrm{i}}}{m}$
$\overrightarrow{a}=\frac{e{E}_0}{m}\hat{\mathrm{i}}$
$V=u+at$
$V=V_0+\frac{e{E}_0}{m}t$
$\because$ De Broglie wavelength$=$ $\lambda =\frac{h}{mv}$ and initially ${\lambda }_0=\frac{h}{m{v}_0}$.
$= \frac{h}{m\left[V_0+\frac{e{E}_{0}t}{m}\right]}=\frac{h}{m{V}_0\left[1+\frac{e{E}_{0}t}{m{V}_0}\right]}=\frac{{\lambda }_0}{1+\frac{e{E}_{0}t}{m{V}_0}}$

Asked in: NEET 2018

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