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A charged particle $q_{0}$ of mass $m_{0}$ is projected along the y-axis at $t = 0$ from origin with velocity $v_{0} .$ There is a uniform electric field $E_{0}$ along x-axis. If the time when de-Broglie wavelength becomes half of initial value is $n \frac{m_{0} v_{0}}{q_{0} E_{0}},$ then find $n^{2} .$

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answer is 3.

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Detailed Solution

$λ = \frac{h}{p}$

$p_{i} = m v_{0}$

To half the wavelength
$p$ has to be doubled
$V_{f} = 2 v_{0} = \sqrt{V_{x}^{2} + V_{y}^{2}}$

$4 V_{0}^{2} = {[\frac{q . E}{m} . t]}^{2} + V_{0}^{2}$

$3 V_{0}^{2} = \frac{q_{0}^{2} . E^{2}}{m^{2}} . t^{2}$

$n = \sqrt{3} \Rightarrow n^{2} = 3$

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